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http://web2.0calc.com/questions/p-p_37645
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+0
# P**p
+5
52
1
x: 0.5(5-7x)=8-(4x+6)
Guest Mar 17, 2017
Sort:
#1
0
Solve for x:
0.5 (5 - 7 x) = 2 - 4 x
Expand out terms of the left hand side:
2.5 - 3.5 x = 2 - 4 x
Add 4 x to both sides:
4 x - 3.5 x + 2.5 = (4 x - 4 x) + 2
4 x - 4 x = 0:
4 x - 3.5 x + 2.5 = 2
4 x - 3.5 x = 0.5 x:
0.5 x + 2.5 = 2
Subtract 2.5 from both sides:
0.5 x + (2.5 - 2.5) = 2 - 2.5
2.5 - 2.5 = 0:
0.5 x = 2 - 2.5
2 - 2.5 = -0.5:
0.5 x = -0.5
Multiply both sides of 0.5 x = -0.5 by 2.:
2.×0.5 x = -0.5×2.
2.×0.5 = 1.:
1. x = -0.5×2.
2. (-0.5) = -1.:
1. x = -1.
Divide both sides by 1.
Guest Mar 17, 2017
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https://www.experts-exchange.com/questions/29115422/T-SQL-solution-to-find-out-how-many-weeks-have-passed-between-2-dates.html
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T-SQL solution to find out how many weeks have passed between 2 dates.
I need a T-SQL solution to find out how many weeks have passed between 2 dates.
The week starts on a Monday and ends on a Friday.
So if the date was this Thursday and the new date is next Monday I need the answer to be 1.
In other words I do not need to be counting full weeks. This is necessary to change dates in multiple weeks of a project when a target date changes.
So if the target date is Thursday 2018/12/06 and it's changed to Tuesday 2018/12/18 my answer would be 2. (remember I do not count full weeks). Each instance passed a Monday is considered a week. And I need it to start changing (adding the weeks) with the current weeks Monday. So if today was Tuesday 11/27/2018 these would be my old dates and new dates. (There are dates before the target date and after the target date).
Original dates. New dates
M-11/26/18:F-11/30/18 M-12/10/18:F-12/14/18
M-12/03/18:F-12/07/18 M-12/17/18:F-12/21/18 Target date
M-12/10/18:F-12/14/18 M-12/24/18:F-12/28/18
M-12/17/18:F-12/21/18 M-12/31/18:F-01/04/19
Who is Participating?
I wear a lot of hats...
"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.
Senior DeveloperCommented:
Well, it's imho simple, depending on the used SQL dialect, e.g. T-SQL:
``````SELECT DATEDIFF(DAY, DateFrom, DateTo) / 7 AS Weeks FROM yourTable;
``````
Associate Principal EngineerCommented:
Hello,
You can try this query
``````declare @from datetime= '9/1/2018'
declare @to datetime = '9/17/2018'
select datediff(day, -7, @to)/7-datediff(day, -6, @from)/7 MondayCount
``````
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Commented:
How many weeks passed two dates:
``````DECLARE @OriginalDate Date = '2018/12/06'
DECLARE @NewDate Date = '2018/12/07'
SELECT DATEDIFF(week, @OriginalDate, @NewDate)
``````
So you can add 7 * DATEDIFF(week, @OriginalDate, @NewDate) to all your existing date values and you should be fine.
Author Commented:
Exactly what I needed.
Thanks!
Commented:
Just to clarify - what should be the result for date change from Monday this week to Friday next week? 1 week or 2 weeks?.
Author Commented:
1 Week
Commented:
Good. The accepted solution calculates this difference as 2 weeks:
``````declare @from datetime= '8/27/2018'
declare @to datetime = '9/7/2018'
select datediff(day, -7, @to)/7-datediff(day, -6, @from)/7 MondayCount
``````
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This is a Clilstore unit. You can .
# Airport: Personal profile – Andrew
We must find a petrol station, but we must plan this carefully and not make any mistakes. We must make a plan of our movements.
We can go north, south, east, or west. We can drive to the right, or left. We can drive back the way we came if we need it.
Fortunately, we have found out that there are signs 80 km before each petrol station and also when we get close to the airport.
We have no proper map of the area. But I have drawn a map with us and the petrol station located in the middle. I have given this square number 1. One square equals one hour of travel. I have also numbered the squares to the north, south, east and west. No number is better than any other, i.e., any number will do. We do know that there is no petrol station in sight from this square, but when we have made move, a sign might turn up.
Which direction should we take? I think we may as well go east. Oh no, wait, we have not been driving in a westerly direction yet. We should try that.
Short url: http://test.multidict.net/cs/938
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https://crypto.stackexchange.com/questions/78212/research-question-part-2-new-symmetric-key-cryptosystem
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# Research Question (Part 2): new symmetric key cryptosystem
A few weeks ago I asked about a new symmetric cryptosystem discovered as consequence of some new mathematics work (link here). As per suggestion of commenters, I spoke with a cryptography professor, a computer security expert, and also continued to discuss with different math professors. We have since established the following facts as given in the below list. We wanted to (for the last time) ask a broad audience that could provide useful feedback given this now established new information.
Encryption takes form $$E(T, K, P) = (T, C)$$, and $$D(T,K, C) = P$$ for publicly chosen $$T$$. Some of the following points have redundancies between each other, but I allow this for clarity. The main properties of the cryptosystem are as follows.
• The security of the system rests on no unproven assumptions (such as the difficulty of factoring).
• Perfect secrecy: for $$x \in P, y \in C$$, $$p(x|y) = p(x)$$ (unconditionally secure), as arbitrary number of ciphertext gives no information about key or message.
• Key reuse: unlike the one-time pad, given arbitrary number of $$(T, C_i)$$ pairs and no plaintext, the attacker learns nothing about $$P_i$$ or $$K$$. Given $$(T, P_i, C_i)$$, the attacker learns nothing about $$P_j$$, $$i \not=j$$, given $$E(T', K, P_j) = (T', C_j)$$ for $$T' \not=T$$, nor about the key $$K$$.
• Secure under chosen plaintext attack: the attacker learns nothing usable about $$K$$ with an arbitrary number of chosen plaintext/ciphertext pairs.
• Semantically secure
• Probabilistic system, but Bob can always successfully decrypt with probability 1
• If the attacker guesses a key $$K$$, they have no way of verifying if the guess is correct, only that it is false.
• If Alice and Bob share $$l$$ messages using the same public object $$T$$, then if Eve obtains a plaintext/ciphertext pair she may decrypt all such messages, but she will be unable to decrypt messages sent using public object $$T' \not=T$$.
• More general cryptanalysis: with large number $$l$$ of distinct $$(T_i, C_i)$$ pairs so that $$(T_i, C_i) \not = (T_j, C_j), i \not=j$$, then, if Eve has a lot of compute, she may brute force construct an approximating key $$K_a$$, that would, with probability proportional to the size of $$l$$, give useful information about any new $$C$$, thereby ruining any previous notion of perfect secrecy with the cipher. However, if Alice and Bob use the same $$T$$ for each message, such an approximating key could not be built, but then the messages would be susceptible to plaintext attack.
• Note on efficiency: worst case encryption and decryption times grow linearly with key size. In most cases encryption/decryption time grows at rate of log of key length.
Again, given this new (verified) information, does this system offer any theoretically interesting properties that could warrant publication? Thank you all for your time.
• I guess the first assumption about it being proven to be secure is the main interesting part. Other than the OTP, I don't think that there are any provable secure schemes out there. If you already have spoken to a number of professors it starts to look interesting, I think they are a better judge if this scheme is worth a paper or not, as we're just being told the properties. Mar 14, 2020 at 23:14
• I am not an expert in details of provably secure systems, though this paper co-authored by a reputable author in the coding theory community claims to have designed one, for whatever it may be worth. I am unfamiliar with the specific journal and did not do any reputation search for the journal: hindawi.com/journals/mpe/2016/7920495 Mar 15, 2020 at 1:30
• I would say that this is not perfect secrecy. Mar 15, 2020 at 16:15
Again, given this new (verified) information, does this system offer any theoretically interesting properties that could warrant publication?
Again, no. You could put together a paper and submit it to eprint (or maybe arxiv, I'm not familiar with their acceptance policies), but beyond that, I can't think of any journal or conference that'd be interested.
As for your "verified" information:
Perfect secrecy: for $$x \in P, y \in C$$, $$p(x|y) = p(x)$$
It is a homework exercise to show that if you have this property, then you must have at least as many keys as possible plaintexts. So, to encrypt a megabyte plaintext, you must have a megabyte key. So, do you have megabyte keys? Or, do you have a low ceiling on the size of the plaintext you can encrypt?
If Alice and Bob share $$l$$ messages using the same public object $$T$$, then if Eve obtains a plaintext/ciphertext pair she may decrypt all such messages
So, the public object $$T$$ is effectively a nonce; that is, a given $$T$$ value can be used to encrypt only a single message.
with large number $$l$$ of distinct $$(T_i, C_i)$$ pairs so that $$(T_i, C_i) \not = (T_j, C_j), i \not=j$$, then, if Eve has a lot of compute, she may brute force construct an approximating key $$K_a$$, that would, with probability proportional to the size of $$l$$, give useful information about any new $$C$$
So, if the attacker gets a number of ciphertexts, they can get information about plaintexts encrypted with the same key. This falls rather short of the "perfect secrecy" you claimed in the previous bullet points.
The only comeback would appear to be "by a lot, you mean more computation than is feasible. However, if you make that argument, you have to assume that there are no optimizations beyond the attack you have sketched out; that means that you are making an assumption (and so you are not really any better than, say, AES or ChaCha…)
worst case encryption and decryption times grow linearly with key size. In most cases encryption/decryption time grows at rate of log of key length.
I'm not sure what to make of this. By 'key length', we typically mean 'length of the key in bits'. To use $$n$$ bits of the key as part of the encryption or decryption time, that takes at least $$O(n)$$ time (if nothing else, to read the key bits). If encryption/decryption time grows (in most cases) logrhythmically with key length, that means in most cases you don't access the majority of the key bits (because you don't have the time). This property would make key searching attacks far more efficient (as the attacker wouldn't have to guess most of the key).
Is that what you really meant?
• Could the downvoter provide the reason? Mar 15, 2020 at 18:19
• I was wondering that myself - do they disagree that there's no venue (other than eprint and arxiv) that would accept it (and if so, could they list one)? Or, do they disagree with my analysis of the various claims that the OP makes? Mar 15, 2020 at 18:41
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Lemma 47.22.4. The following types of rings have a dualizing complex:
1. fields,
2. Noetherian complete local rings,
3. $\mathbf{Z}$,
4. Dedekind domains,
5. any ring which is obtained from one of the rings above by taking an algebra essentially of finite type, or by taking an ideal-adic completion, or by taking a henselization, or by taking a strict henselization.
Proof. Part (5) follows from Proposition 47.15.11 and Lemma 47.22.3. By Lemma 47.21.3 a regular local ring has a dualizing complex. A complete Noetherian local ring is the quotient of a regular local ring by the Cohen structure theorem (Algebra, Theorem 10.160.8). Let $A$ be a Dedekind domain. Then every ideal $I$ is a finite projective $A$-module (follows from Algebra, Lemma 10.78.2 and the fact that the local rings of $A$ are discrete valuation ring and hence PIDs). Thus every $A$-module has finite injective dimension at most $1$ by More on Algebra, Lemma 15.69.2. It follows easily that $A[0]$ is a dualizing complex. $\square$
Comment #1684 by on
It's perhaps better to say that (5) follows from 0A7K in general. In the current proof there's just "any ring essentially of finite type over a regular local ring [is Gorenstein]". This is a bit confusing.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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# A whole number or the quotient of any whole numbers, excluding zero as a denominator A whole number or the quotient of any whole numbers, excluding zero.
## Presentation on theme: "A whole number or the quotient of any whole numbers, excluding zero as a denominator A whole number or the quotient of any whole numbers, excluding zero."— Presentation transcript:
A whole number or the quotient of any whole numbers, excluding zero as a denominator A whole number or the quotient of any whole numbers, excluding zero as a denominator Examples - 5/8; -3/14; 7/-15; -6/-11 Examples - 5/8; -3/14; 7/-15; -6/-11 Natural numbers – They are counting numbers. Natural numbers – They are counting numbers. Integers - Natural numbers, their negative and 0 form the system of integers. Integers - Natural numbers, their negative and 0 form the system of integers. Fractional numbers – the positive integer which are in the form of p/q where q is not equal to 0 are known as fractional numbers. Fractional numbers – the positive integer which are in the form of p/q where q is not equal to 0 are known as fractional numbers.
Closure Property Rational numbers are closed under addition. That is, for any two rational numbers a and b, a+b s also a rational number. Rational numbers are closed under addition. That is, for any two rational numbers a and b, a+b s also a rational number. For Example - 8 + 3 = 11 ( a rational number. ) Rational numbers are closed under subtraction. That is, for any two rational numbers a and b, a – b is also a rational number, For Example - 25 – 11 = 14 ( a rational number. ) Rational numbers are closed under multiplication. That is, for any two rational numbers a and b, a * b is also a rational number. For Example - 4 * 2 = 8 (a rational number. ) Rational numbers are not closed under division. That is, for any rational number a, a/0 is not defined. For Example - 6/0 is not defined.
Commutative Property Rational numbers can be added in any order. Therefore, addition is commutative for rational numbers. For Example – Subtraction is not commutative for rational numbers. For Example - Since, -7 is unequal to 7 Hence, L.H.S. Is unequal to R.H.S. Therefore, it is proved that subtraction is not commutative for rational numbers. L.H.S. R.H.S. - 3/8 + 1/7 L.C.M. = 56 = -21+8 = -13 1 /7 +(-3/8) L.C.M. = 56 = 8+(-21) = -13 L.H.S. R.H.S. 2/3 – 5/4 L.C.M. = 12 = 8 – 15 = -7 5/4 – 2/3 L.C.M. = 12 = 15 – 8 = 7
Rational numbers can be multiplied in any order. Therefore, it is said that multiplication is commutative for rational numbers. For Example – Since, L.H.S = R.H.S. Therefore, it is proved that rational numbers can be multiplied in any order. Rational numbers can not be divided in any order.Therefore,division is not commutative for rational numbers. For Example – Since, L.H.S. is not equal to R.H.S. Therefore, it is proved that rational numbers can not be divided in any order. L.H.S. R.H.S. -7/3*6/5 = -42/15 6/5*(7/3) = -42/15 L.H.S. R.H.S. (-5/4) / 3/7 = -5/4*7/3 = -35/12 3/7 / (-5/4) = 3/7*4/-5 = -12/35
Associative property Addition is associative for rational numbers. Addition is associative for rational numbers. That is for any three rational numbers a, b and c, a + (b + c) = (a + b) + c. That is for any three rational numbers a, b and c, a + (b + c) = (a + b) + c. For Example For Example Since, -9/10 = -9/10 Hence, L.H.S. = R.H.S. Therefore, the property has been proved. has been proved. Subtraction is not associative for rational numbers. For Example - Since, 19/30 is not equal to 29/30 Hence, L.H.S. is not equal to R.H.S. Therefore, the property has been proved. L.H.S. L.H.S. R.H.S. R.H.S. -2/3+[3/5+(-5/6)] = -2/3+(-7/30) = -27/30 = -9/10 [-2/3+3/5]+(-5/6)=-1/15+(-5/6)=-27/30=-9/10 -2/3-[-4/5-1/2] -2/3-[-4/5-1/2] = -2/3 + 13/10 =-20 +39 /30 = 19/30 [2/3-(-4/5)]-1/2 = 22/15 – ½ = 44 – 15/30 = 29/30 = 29/30
Multiplication is associative for rational numbers. That is for any rational numbers a, b and c a* (b*c) = (a*b) * c For Example – Since, -5/21 = -5/21 Hence, L.H.S. = R.H.S Division is not associative for Rational numbers. for Rational numbers. For Example – Since, Hence, L.H.S. Is Not equal to R.H.S. equal to R.H.S. L.H.S. L.H.S. R.H.S. R.H.S. -2/3* (5/4*2/7) = -2/3 * 10/28 = -2/3 * 5/14 = -10/42 = -5/21 (-2/3*5/4) * 2/7 = -10/12 * 2/7 = -5/6 * 2/7 = -10/42 = -5/21 L.H.S. L.H.S. R.H.S. R.H.S. ½ / (-1/3 / 2/5) = ½ / -5/6 = -6/10 = -3/5 [½ / (-1/2)] / 2/5 = -1 / 2/5 = -5/2
Distributive Law Distributivity of multiplication over addition and subtraction : Distributivity of multiplication over addition and subtraction : For all rational numbers a, b and c, For all rational numbers a, b and c, a (b+c) = ab + ac a (b+c) = ab + ac a (b-c) = ab – ac a (b-c) = ab – ac For Example – For Example – Since, L.H.S. = R.H.S. Hence, distributive law is proved L.H.S. L.H.S. R.H.S. R.H.S. 4 (2+6) 4 (2+6) = 4 (8) = 32 4*2 + 4*6 4*2 + 4*6 = 8 = 24 = 32
The Role Of Zero (0) Zero is called the identity for the addition of rational numbers. It is the additive identity for integers and whole numbers as well. Zero is called the identity for the addition of rational numbers. It is the additive identity for integers and whole numbers as well. Therefore, for any rational number a, a+0 = 0+a = a Therefore, for any rational number a, a+0 = 0+a = a For Example - 2+0 = 0+2 = 2 For Example - 2+0 = 0+2 = 2 -5+0 = 0+(-5) = -5 -5+0 = 0+(-5) = -5 The role of one (1) The role of one (1) 1 is the multiplicative identity for rational numbers. 1 is the multiplicative identity for rational numbers. Therefore, a*1 = 1*a = a for any rational number a. Therefore, a*1 = 1*a = a for any rational number a. For Example - 2*1 = 2 For Example - 2*1 = 2 1*-10 = -10 1*-10 = -10
Additive Inverse Additive inverse is also known as negative of a number. For any rational number a/b, a/b+(-a/b)= (-a/b)+a/b = 0 Therefore, -a/b is the additive inverse of a/b and a/b is the additive inverse of (-a/b) Reciprocal Rational number c/d is called the reciprocal or multiplicative inverse of another rational number a/b if a/b * c/d = 1
Some Problems On Rational Numbers Q 1 ) Verify that –(-x) is the same as x for x = 5/6 A 1 ) The additive inverse of x = 5/6 = -x = -5/6 of x = 5/6 = -x = -5/6 Since, 5/6 + (-5/6) = 0 Since, 5/6 + (-5/6) = 0 Hence, -(-x) = x. Hence, -(-x) = x. Q 2 ) Find any four rational numbers between -5/6 and 5/8 A 2 ) Convert the given numbers to rational numbers with same denominators : -5*4-6*4 = -20/24 5*3/8*3 = 15/24 -5*4-6*4 = -20/24 5*3/8*3 = 15/24 Thus, we have -19/24; -18/24;........13/24; 14/24 Thus, we have -19/24; -18/24;........13/24; 14/24 Any four rational numbers can be chosen. Any four rational numbers can be chosen. L.H.S. L.H.S. R.H.S. R.H.S. -(-5/6) = +5/6 = 5/6 5/6 = + 5/6 = 5/6
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# Use Gauss's law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities σ and −σ respectively.
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## Electric field due to a uniformly charged infinite plane sheet : Suppose a thin non-conducting uniform surface charge density σ. Electric field intensity →E on either side of the sheet must be perpendicular to the plane of sheet having same magnitude at all points equidistant from sheet. Let P be any point at a distance r from the sheet. Let the small area element −→dS=dSˆn. →E and ˆn are perpendicular on the surface of imagined cylinder, so electric flux is zero. →E and ˆn are parallel on the two cylindrical edges P and Q which contributes electric flux. ∴ Electric flux over the edges Pand Q of the cylinder is 2ϕ=qε0⇒2∮→E−→dS.=qε0 ⇒2∮→E−→dS.=qε0 ⇒2∮EdS=qε0⇒2Eπr2=qε0⇒E=q2πε0r2 ∴ The charge density σ=qS ⇒q=πr2σ E=πr2σ2πε0r2 E=σ2ε0, vectorically →E=σ2ε0ˆn Where ˆn is a unit vector normal to the plane and going away from it. where σ>0 E is directed away from both sides. Now consider two infinite plane parallel sheets of charge A and B. Let σ1=σ and σ2=−σ be the uniform surface density of charge on A and B respectively. The electric field between two plates is given by E=E1−E2 =σ12ε0−σ22ε0=σ2ε0−(−σ2ε0)=σ+σ2ε0 ∴E=2σ2ε0⇒E=σε0.
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Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
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Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
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Author Message
DK
Registered: 31.08.2001
From: Champaign, IL
Posted: Thursday 28th of Dec 09:36 Hey Friends , I am urgently in need of guidance for passing my math exam that is approaching . I really do not want to opt for the assistance of private masters and web coaching because they prove to be quite costly. Could you recommend a perfect tutoring software that can help me with learning the intricacies of Algebra 2. Particularly, I need assistance on mixed numbers and graphing circles.
ameich
Registered: 21.03.2005
From: Prague, Czech Republic
Posted: Friday 29th of Dec 10:51 There used to be a time when those were the only ways out. But thanks to technology now, we have something known as the Algebrator! It’s an easy to use software , something which even a total newbie would enjoy working on and the best part is that it would solve all your questions and also explain the steps it took to reach that solution ! Isn’t that fabulous? Well I think it is, and I’m sure after using it trying it, you won’t disagree with me .
DVH
Registered: 20.12.2001
From:
Posted: Saturday 30th of Dec 07:26 I agree. Stress will lead you to doom . Algebrator is a very handy tool. You don’t need to be a computer pro in order to use it. Its easy to use, and it works great.
jwumdirlj
Registered: 30.03.2003
From: Easton, PA (in the groove)
Posted: Sunday 31st of Dec 11:16 Wow, sounds really good! Can you tell me where I can get more details? I would like to purchase a copy of this software immediately.
SjberAliem
Registered: 06.03.2002
From: Macintosh HD
Posted: Tuesday 02nd of Jan 11:32 Algebrator is the program that I have used through several algebra classes - Basic Math, Pre Algebra and Intermediate algebra. It is a truly a great piece of algebra software. I remember of going through difficulties with converting fractions, decimals and angle-angle similarity. I would simply type in a problem homework, click on Solve – and step by step solution to my math homework. I highly recommend the program.
Dolknankey
Registered: 24.10.2003
From: Where the trout streams flow and the air is nice
Posted: Thursday 04th of Jan 11:28 Is that so? It is uncomplicated to acquire this program. In addition , you have nothing to lose. The program comes with a money-back guarantee . It is available at: http://www.algebra-expression.com/rational-expressions-1.html. I am confident that you will simply enjoy it. Let me know if there is something more that you would like to know from me.
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# Function Calculator
## Calculate the properties of a function step by step
The calculator will try to find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical (stationary) points, extrema (minimum and maximum, local, relative, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single-variable function. The interval can be specified. Parity will also be determined.
Enter a function of one variable:
Enter an interval:
Required only for trigonometric functions. For example, [0, 2pi] or (-pi, oo). If you need oo, type inf.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
The Function Calculator is a tool that allows you to many properties of functions. Easily explore functions by examining their parity, domain, range, intercepts, critical points, intervals of increase/decrease, local and global extrema, concavity intervals, inflection points, derivatives, integrals, asymptotes, and so on.
## How to Use the Function Calculator?
• ### Input
Enter the function you want to analyze. If necessary, enter the interval which you are interested in.
• ### Calculation
Click the "Calculate" button to start the process.
• ### Result
The calculator will quickly display the properties of the function.
## What Is a Function?
In mathematics, a function is a formula that, for every value from the input set (domain), produces some value. The set of such values forms the output set (range, codomain). The function is called one-to-one if no two elements from its domain give the same element from the range.
Functions are typically denoted using various notations. The most common notation is $f(x)$. In the notation $f(x)$ $f$ is the function symbol and $x$ represents the input variable.
For example, $f(x)=x^2$ is a quadratic function. If $x=-2$, then $f(-2)=(-2)^2=4$. If $x=2$, then $f(2)=2^2=4$. By the way, it is not one-to-one, since there are two elements, namely $-2$ and $2$, that give $4$.
## What Are the Components of a Function?
• Domain: The domain of a function is the set of all possible input values. It defines where the function is defined or valid.
• Codomain: The codomain of a function is the set of all possible output values. It represents the values that the function can produce.
• Range: The range of a function is the set of all actual output values. It represents the values that the function actually produces.
• Input: The input of a function is the value of its variable you provide as an argument to the function. It's the value for which you want to determine the corresponding output.
• Output: The output of a function is the result of applying the function to the input value(s). It represents the dependent variable's value corresponding to the given input.
## What Is the Mathematical Expression of a Function?
A function is often expressed using a formula or rule that describes how the input values are transformed into output values. This formula can be very simple, for example, $f(x)=x$, or very complex, for example, $f(x,y,z)=\frac{x^2+y^2z}{z-xy}$ (a multivariable function). Here are some examples:
• Linear Function. A linear function has the following form:
$$f(x)=mx+b$$
Example: $f(x)=2x+3$.
This function represents a straight line on a graph, where $m$ is the slope, and $b$ is the y-intercept.
$$f(x)=ax^2+bx+c$$
Example: $f(x)=3x^2-4x+1$.
This function represents a parabolic curve.
• Exponential Function. An exponential function has the following form:
$$f(x)=a\cdot b^x$$
Example: $f(x)=2\cdot3^x$.
This function exhibits exponential growth or decay.
## Why Choose Our Function Calculator?
• ### Ease of Use
Our calculator is designed with user-friendliness in mind. It features an intuitive interface that allows both beginners and experts to navigate with ease.
• ### Versatility
Our calculator can handle a wide range of functions to meet your requirements.
• ### Accurate Results
Our calculator provides accurate results. You can trust it.
### FAQ
#### What is meant by the graph of a function?
The graph of a function is a visual representation that shows how the output (dependent variable) of the function relates to the input (independent variable). It is a plot of points where each point corresponds to an input-output pair of values. The graph illustrates the behavior of a function—its shape, intersection points, critical points, inflection points, and so on.
#### What is the difference between a relation and a function?
A relation is a set of ordered pairs $(x,y)$, where $x$ is the input (independent variable), and $y$ is the output (dependent variable). It can include multiple y-values for the same x-value. In contrast, a function is a special type of relation, which is defined by some rule, so you can calculate the output value, given the input value.
#### What is the Function Calculator used for?
The Function Calculator is a tool used to analyze functions. It can find the following for a function: parity, domain, range, intercepts, critical points, intervals of increase/decrease, local and global extrema, concavity intervals, inflection points, derivative, integral, asymptotes, and limit. The calculator will also plot the function's graph.
#### How accurate are the results from this calculator?
You can trust our calculator because it provides accurate results.
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Reader Pattu  has developed a retirement calculator for his own use, and was interested in sharing it with other OneMint readers as well. I took a look at it, and liked what he’s done, so I’m sharing the calculator here.
He has also described his thought process, which I’m sharing below.
Retirement planning is a complicated financial goal. When you save for a new car, exotic holiday or even for the education of your children, all or most of the accumulated corpus would get spent when the time for the goal arrives. So it is easier to calculate the corpus required for such goals.
When it comes to retirement planning the corpus calculation is complicated because the corpus does not get spent in one shot. Typically it is allowed to grow at some post-retirement interest rate (usually underestimated for safety) and monthly withdrawals are made from it. For complete financial independence during retirement, these withdrawals or pension must increase according to the post-retirement inflation rate.
So the retirement corpus calculation has to take into account not only the years to retirement, present inflation rate but also post-retirement inflation, number of years in retirement and post-retirement interest rate on the corpus.
Given these input parameters, retirement planning consists of two major steps.
1. Calculation of the corpus required:
(a) Compute projected future monthly expenses at the start of retirement. This amount can be taken as the initial monthly pension which will be withdrawn from the corpus.
(b) This monthly pension amount is assumed to increase every year according to the rate of inflation, while the corpus is assumed to grow at some constant interest rate. Using these inputs the corpus is calculated using Excel’s present value calculator (PV). The inputs to the PV function, the formulae used are a bit technical and can be found here.
(The site also has a nice annuity calculator which will be of interest to people close to retirement).
Since the withdrawals are indexed to inflation, the corpus will decrease each year and become zero at the end of the retirement period given as input. A typical rise and fall of the corpus is shown here.
2. Calculation of the monthly investment amount required:
Once the corpus needed is known this calculation is similar to any other financial goal. It uses the number of years to retirement, estimates of present inflation and annual interest rate, annual increase in investment if any, and amount accumulated so far if any.
The excel based calculator does the above tasks and present an annual cash flow chart of the monthly investment made, monthly expenses, the growth of the retirement corpus prior to retirement and its decrease post retirement. Annual salary is also shown for reference.
1. The calculator has been repeatedly tested and checked. However no guarantee is made that it is free of errors. Like every other tool it serves only as an estimate for the monthly investment required and depends on the input parameters. It is meant to be used for education purposes only.
2. For simplicity the pre- and post-retirement inflation rates are taken to be the same. Use a reasonably high value to be safe. A financial expert suggested using 8-9%
3. The pre-retirement interest rate is taken as constant. Since equity allocation would decrease as retirement approaches this would change. The calculator can easily be modified to take into account variable interest rate. Contact me if you need to look at such a calculation.
Bugs, comments and suggestions are welcome.
Pattu
pattu@iitm.ac.in
Update: Fixed a divide by zero error bug.
## 54 thoughts on “Download Excel Based Retirement Calculator”
1. narayan meher says:
Rd scheem ROI details of every month
2. Dear Sir,
Nice work on pension calculator.
3. GK says:
Just downloaded the looks really good…will enter the details and test it.
4. Allen Maloney says:
Very impressive work. I am grateful.
Any possibility to two additional incomes that are differed in time.
Per example:
1- A pension that would start at 70; and
2- Another fixed income that would be come at another age.
Making them subject each to an individual inflation rate would be fantastic.
Thanks in advance for considering, whether or not you do it.
1. Allen Maloney, these options have been incorporated in different forms in my website.
5. Excellent post. I certainly appreciate this site.
Stick with it!
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basic lotion again, to the areas mentioned
before. It is preferable to use a moisturizer that is meant for a sunless tanned skin.
7. 100.00 to 100000.00 recaring deposite rate of interist 1 month to 10 year . in pdf file
8. 100 to 100000 rate of intarest monthely calculation in 1 month to 10 year . pdf file
9. J S Jolly says:
Dear Pattu
I was trying to locate such an excel calculator for quite some time. Excellent. You have made my pension calculations, very easy.
1. Thanks. You could also check out my other retirement calculators and investing calculators in my blog (freefincal.com)
10. Suraj says:
Hi Pattu,
Shouldn’t your corpus get exhausted within “Estimated years of retirement” ?
I did some runs for 80 year life span and saw some corpus left so extended the life span and saw the same corpus sustaining me till 100 !
1. Hi Suraj,
Yes it should. Can you send me the file so that I can have a look? If you get some time, check my other retirement calculators and see how they fare.
Thanks
Pattu
11. Anand K says:
Great excel too. Grateful to u to have shared. Tweaked the parameters to estimate my Term Plan insurance amount also. 🙂
1. Hi Anand,
That is interesting. Could I have a look at what you did?
Btw I have more retirement, insurance and other calculators at
freefincal.wordpress.com
1. Yeah, if possible, I’d like to see it too. Thank you.
12. Santhosh says:
i am quite opposed to equities being compounded annually.It isnt simple as that.It undermines the total compounding effect.debt too cant be compounded annually.Anyway it always gives a measure of how your money grows.
1. Understand your point. The alternative is the problem. One can either use Monte Carlo calculators (I have one in my site which you can try) or to use a return lower than long term historical sensex returns. Nothing is perfect though.
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# Existence of solutions of polynomials systems (and their "rough" shape) over $\mathbb{R}$ & friends with positive-dimensional ideals
This is a follow-up (but self-contained) question to my previous one. There I asked about state-of-the-art methods to solve multivariate polynomials systems over non-algebraically closed fields in general.
I learned that the theory is more involved that I thought (I'm not working in algorithmic algebraic geometry, so I'm only familiar with the very basics, like Buchberger's algorithm, or the definition of the dimension of an ideal). Therefore it is necessary to ask a more specific questions that the previous, general one, which is more tailored to my needs.
My setup is the following:
Regarding complexity: I'm interested in solving a large number of polynomial systems (on commodity hardware), on the order of $$10^4$$. But each of the systems is of relatively small size - my baseline consists of least 6 different variable and 4 equations. If I could tackle this, I'd already be happy. Going further, I don't expect the systems to grow beyond about 20 different variables and 20 equations.
So perhaps I don't actually need the fastest possible algorithm and can make do with simpler, older ones - but I will let you be the judge of that.
Regarding the polynomials: There are no restrictions their coefficients, so, depending on the field I'm working in they can take any number.
Regarding the field: Regarding the field I'm working in, my baseline is $$\mathbb{R}$$, but I'd also be interested in $$\mathbb{Q}$$ and $$\mathbb{Z}$$. If there are methods that are much easier for one field than another, than I will the choice of the field to study be influence by the time I need to invest to learn that method, i.e. the easiest one wins.
Regardin the dimension of the ideal spanned by the polynomials: The ideal has dimension $$2$$ or $$3$$ over the complex numbers, in most cases I tested so far with the help of CAS.
What I'm looking for: I'm interested is learning about methods (I'm happy with specific references) that tell me
1) whether the system has a solution at all or not. Working over, e.g., $$\mathbb{C}$$, this would be easy (e.g. compute a Gröbner basis: If it contain the $$1$$, if and only if the solution variety is empty). But this doesn't work unfortunately for non-algebraically closed fields. Given the answers from my previous question, I'm inclined to think that answer this question shouldn't be too hard (perhaps even trivial for the expert computational geometer, which I'm not unfortunately).
2) if it has an infinite number of solutions (if the variety is zero-dimensional, things are easy of course), I would like to pick out one single variable, say $$n_0\in \{1,\ldots,n\}$$, project the solution variety $$V(f_1,\ldots,f_s)\subseteq \mathbb{ R}^n$$ (supposing we work over the field $$\mathbb{R}$$) along this variable onto $$\mathbb{R}$$ to investigate whether there exists an interval $$[-\alpha,\alpha]$$ around $$0$$ which is contained in this projected set (I don't need to understand the projected set fully). That is what I menat by "rough shape" in the title.
• Are the coefficients exact or only floating-point numbers? May 29, 2020 at 8:37
• @FrançoisBrunault My investigation has two steps: First, experiments, where I necessarily use floating-point precision. In a second step, based on the experiments, I will want to prove some general statements, where I will allow any real-numbers as coefficients (even if these can't be simulated any more). I'm guessing there seems to be a dramatic difference in terms of the capability of the methods you have in mind between these two cases. While I, for now, would be interested mostly in the floating-point case to do the experiments, it would be good to know if there are [...] May 29, 2020 at 9:11
• some fundamental issues I might miss when doing experiments, which come from floating points. Thus I would ask you to focus on floating-point, but dt mention where I need to be careful, when passing to exact real coefficients (so that I will know where any pitfalls might lie when generalizing after I have finished experimenting; perhaps; if the exact coefficient case is more difficult I could adjust my investigation accordingly). May 29, 2020 at 9:14
• If the coefficients are floating point then your question is somewhat ill-posed, because polynomial equation solving is very sensitive to the input. Think of the equation $x^2=\varepsilon$ with $\varepsilon$ near 0, the answer to your question (2) changes completely. You can certainly approximate the coefficients by decimal numbers, and apply the quantifier elimination method I mentioned to get the projection as an explicit semi-algebraic set of $\mathbb{R}$. And it's possible to decide whether a semi-algebraic set is a neighbourhood of $0$. But you have no guarantee that the answer is correct May 29, 2020 at 9:20
• @FrançoisBrunault Well, by necessity I need to use floating point, since I don't know of a different way to carry out solving the mentioned $10^4$ systems, where the coefficients discretize a high-dimensional cube (though, afterwards, when I want to prove theorems, which is the final goal, of course I will work with coefficient from $\mathbb{R}$). May 29, 2020 at 9:38
I assume that your polynomials have rational coefficients (which seem to be the case, since you mention they are floating point numbers with fixed precision, in particular they are decimals), and that you are interested in the solutions in $$\mathbb{R}^n$$.
The assertion that the projection of $$V(f_1,\ldots,f_s) \subset \mathbb{R}^n$$ to the $$x_n$$ variable is a neighbourhood of 0 is a first order formula over the reals, namely $$\begin{equation*} \exists a>0, \forall x_n \in [-a, a], \exists x_1,\ldots,x_{n-1} \in \mathbb{R}, \forall i, f_i(x_1,\ldots,x_n) = 0. \end{equation*}$$ It is a formula with no free variable, hence is decidable, and CAD softwares like Qepcad or Redlog will output "true" or "false".
Regarding feasibility, my worry is that the semi-algebraic set of $$\mathbb{R}$$ given by the projection to $$x_n$$ will probably involve polynomials with gigantic coefficients. You have to experiment to see whether the CAD software can still do it in reasonable time.
• @user43263 By the way, I realize that this also works (in theory) for polynomials whose coefficients are arbitrary real numbers (which are computable to arbitrary precision, like $\pi$, $e$ or whatever...). Namely, you introduce indeterminates for these coefficients. CAD will output a boolean formula involving (in)equalities between polynomials in these indeterminates (like "$b^2-4ac \geq 0$"), and you just need to evaluate these expressions to sufficient precision to decide whether the boolean formula is true or false. May 29, 2020 at 13:23
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# Reflexive Property
Written by Jerry Ratzlaff on . Posted in Algebra
A postulate is a statement that is assumed true without proof.
Let $$\;a\;$$ , $$\;b\;$$ and $$\;c\;$$ be real numbers.
• $$\;a = a\;$$ (A quantity is congruent (equal) to itself.)
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+0
# Trig Problems
0
99
1
+23
prove these identities without interacting across the equals sign (i.e. subtracting from both sides or cross multiplyin)
1.
1/(csc x + cot x) = (1-cos x)/sin x
2.
sec x/(1+cos x) = csc^2(x) (sec x -1)
Apr 22, 2021
#1
+420
+2
1.
$$\frac{1}{\csc x + \cot x}\\ =\frac{1}{\frac{1}{\sin x} + \frac{\cos x }{\sin x}}\\ =\frac{1}{\frac{1+\cos x}{\sin x}}\\ =\frac{\sin x}{1+ \cos x}\\ = \frac{(\sin x)(1- \cos x)}{(1+\cos x)(1- \cos x)}\\ = \frac{(\sin x)(1- \cos x)}{1-(\cos x)^2}$$
by the Pythagorean identity, $$(\sin x)^2 + (\cos x)^2 = 1$$ (you could prove using Pythagorean's theorem), which means that $$1-(\cos x)^2=(\sin x)^2$$. Substitute that back in:
$$\frac{(\sin x)(1-\cos x)}{(\sin x)(\sin(x)}=\frac{1-\cos x}{\sin x}$$ and we are done.
2.
$$(\csc x)^2(\sec x-1)\\ =(\frac{1}{(\sin x)^2})(\frac{1}{\cos x}-1)$$
Use the Pythagorean identity again:
$$(\frac{1}{(1+\cos x)(1-\cos x)})(\frac{1-\cos x}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\frac{1}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\sec x)\\ =\frac{\sec x}{1+\cos x}$$
and we are done.
Apr 22, 2021
edited by textot Apr 22, 2021
#1
+420
+2
1.
$$\frac{1}{\csc x + \cot x}\\ =\frac{1}{\frac{1}{\sin x} + \frac{\cos x }{\sin x}}\\ =\frac{1}{\frac{1+\cos x}{\sin x}}\\ =\frac{\sin x}{1+ \cos x}\\ = \frac{(\sin x)(1- \cos x)}{(1+\cos x)(1- \cos x)}\\ = \frac{(\sin x)(1- \cos x)}{1-(\cos x)^2}$$
by the Pythagorean identity, $$(\sin x)^2 + (\cos x)^2 = 1$$ (you could prove using Pythagorean's theorem), which means that $$1-(\cos x)^2=(\sin x)^2$$. Substitute that back in:
$$\frac{(\sin x)(1-\cos x)}{(\sin x)(\sin(x)}=\frac{1-\cos x}{\sin x}$$ and we are done.
2.
$$(\csc x)^2(\sec x-1)\\ =(\frac{1}{(\sin x)^2})(\frac{1}{\cos x}-1)$$
Use the Pythagorean identity again:
$$(\frac{1}{(1+\cos x)(1-\cos x)})(\frac{1-\cos x}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\frac{1}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\sec x)\\ =\frac{\sec x}{1+\cos x}$$
and we are done.
textot Apr 22, 2021
edited by textot Apr 22, 2021
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http://gamedev.stackexchange.com/questions/13326/how-do-i-generate-projectiles-toward-the-mouse-pointer
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# How do I generate projectiles toward the mouse pointer? [duplicate]
I'm making a top-down space shooter where the player controls a ship and can aim and shoot using the mouse cursor. How can I fire bullets from the ship at the angle from the ship to the mouse cursor?
-
## marked as duplicate by Josh Petrie♦Aug 10 '14 at 4:33
Is it that you need help calculating the angle based on the mouse cursor? – Nate Jun 7 '11 at 20:41
Yeah, sorry if my question is somewhat vague in that regard. And also how to move the projectile based on the angle, that too I need. – Tristan Dube Jun 7 '11 at 21:42
Wait, you're asking us about basic trigonometry, or the means to implement it? – jcora Jan 21 '12 at 20:45
If I understood your problem properly, you just want to shoot a bullet towards a mouse position. Here is how I would do:
First of all, you must find the movement required for the bullet to get to the mouse, like so:
``````Vector2 movement = mousePosition - bulletStartPosition;
``````
Then, you should normalize it to have a vector with a length of 1 so that you can hold a vector which tells you in which direction to go, like so:
``````movement.Normalize();
``````
But here you have a little problem, if the direction is equal to `(0, 0)` (meaning that the mouse is on the bullet start position), then you'll divide by zero, so make sure you check for that with the last piece of code:
``````if (movement != Vector2.Zero)
movement.Normalize();
``````
So, you've got the movement required to move towards the mouse. You have to keep a `Vector2` within your bullet class which holds the `Direction` of the bullet.
What's next? You have to actually move the bullet!
In your bullet update code, do the following:
``````bullet.Position += bullet.Direction * bullet.Speed * gameTime.ElapsedGameTime.TotalSeconds; // multiply by delta seconds to keep a consistent speed on all computers.
``````
Where `bullet.Speed` is a float representing the bullet's speed in units per second.
Basically, here are the things to change:
Inside you bullet class, add a `float Speed` and a `Vector2 Direction`.
When shooting, set your `bullet.Direction` to `mousePosition - bullet.Position` and safely normalize it (by checking for equality with `Vector2.Zero` first).
When updating your bullet, do the following: `bullet.Position += bullet.Direction * bullet.Speed * gameTime.ElapsedGameTime.TotalSeconds;`.
It should work.
-
+1 Vectors are a wonderful thing. Learn them! – doppelgreener Jun 7 '11 at 22:43
Thanks, that did the trick ! Awesome link on the vectors too, definitely something I'll read soon. – Tristan Dube Jun 8 '11 at 1:27
what is movement used for after you normalize it? I dont see it being used. – Raptrex Aug 23 '11 at 18:40
@Raptrex: Actually movement is what I called the Direction later on. I just called it `movement` in the code to first show what it represents. It is basically just the Direction vector. – Jesse Emond Aug 28 '11 at 2:13
@JesseEmond Welcome to the vote to close privilege, use it wisely. – Byte56 May 17 '13 at 19:44
If you're having trouble calculating the angle, you can use this:
``````Vector2 target = mousePos - startPos;
float angle = Math.Atan2( target.Y, target.X );
``````
-
+1 Keepin' it simple. – Arcane Engineer Jan 21 '12 at 16:40
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NCL Home > Documentation > Functions > Meteorology
# tlcl_td_bolton
Compute the lifting condensation level temperature using dew point temperature.
Available in version 6.5.0 and later.
## Prototype
```load "\$NCARG_ROOT/lib/ncarg/nclscripts/csm/contributed.ncl" ; This library is automatically loaded
; from NCL V6.2.0 onward.
; No need for user to explicitly load.
function tlcl_td_bolton (
t : numeric, ; float, double, integer only
td : numeric,
iounit [2] : integer
)
return_val [dimsizes(t)] : float or double
```
## Arguments
t
A scalar or array containing temperature (units: degC, degK, degF). If t and td are arrays they must be the same size and shape.
td
A scalar or array containing dew point temperatures with the same units as t. If td and t are arrays they must be the same size and shape.
iounit
An integer array of length 2 which specifies the units of the input t and the units of the returned variable.
• iounit(0)=0 input t and td are degrees Celcius (degC)
• iounit(0)=1 input t and td are degrees Kelvin (degK)
• iounit(0)=2 input t and td are degrees Farenheit (degF)
• iounit(1)=0 output tlcl are degrees Celcius (degC)
• iounit(1)=1 output tlcl are degrees Kelvin (degK)
• iounit(1)=2 output tlcl are degrees Farenheit (degF)
## Return value
A variable of the same size and shape as t.
## Description
The lifted condensation level or lifting condensation level (LCL) is formally defined as the height or pressure at which the relative humidity (RH) of an air parcel will reach 100% with respect to liquid water when it is cooled by dry adiabatic lifting. The temperature at the LCL is denoted as the 'lifted condensation level temperature' (tlcl). In theory, the lifting condensation level (LCL) is for a parcel of air lifted from the surface.
This function is based on Equation 15 in Bolton (1980).
An interactive lifting condensation level calculator is available here.
```References:
Bolton, D. (1980): The Computation of Equivalent Potential Temperature
Monthly Weather Review, vol. 108, no. 7 (july), p. 1047
Wikipedia: Lifted Condensation Level
```
## Examples
Example 1: Calculate the temperature at the lifted condensation level (tlcl) using tlcl_td_bolton. The online site: Calculating LCL Temperature gives the following result: tlcl=8.8765C.
```
t = 20 ; C ==> iounit(0)=0
td = 13.5 ; C
tlcl = tlcl_td_bolton(t, td, (/0,0/)) ; tlcl=8.8766 ; C ==> iounit(1)=0
printVarSummary(tlcl)
```
The variable output which contains meta data is:
```
Variable: tlcl
Type: float
Total Size: 4 bytes
1 values
Number of Dimensions: 1
Dimensions and sizes: [1]
Coordinates:
Number Of Attributes: 4
long_name : temperature: LCL
units : degC
source : Bolton (1980): Eq. 15: Dew Point Temperature
NCL : tlcl_td_bolton
8.876617
Example 2: Read temperature (T [degK]) and relative humidity (RELHUM [%]); (a) compute the dew point temperatures;
(b) calculate the temperatures at the lifted condensation level.
t = a->T ; K ==> iounit(0)=1
rh = a->RELHUM ; %
td = dewtemp_trh(t ,rh) ; dewtemp_trh requires ttlcl_td_bolton(t, td, (/1,1/)) ; K ==> iounit(1)=1
printVarSummary(tlcl)
printMinMax(tlcl,0)
The output is:
Variable: tlcl
Type: float
Total Size: 1437696 bytes
359424 values
Number of Dimensions: 4
Dimensions and sizes: [time | 1] x [lev | 26] x [lat | 96] x [lon | 144]
Coordinates:
time: [3132..3132]
lev: [3.54463800000001..992.5560999999998]
lat: [ -90..89.99999999999999]
lon: [ 0..357.5]
Number Of Attributes: 4
long_name : temperature: LCL
units : degK
source : Bolton (1980): Eq. 15: Dew Point Temperature
NCL : tlcl_td_bolton
temperature: LCL (degK) : min=164.972 max=298.26
```
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# Vectors
• Aug 7th 2013, 08:11 AM
Kristen111111111111111111
Vectors
(alpha)x + a × x = b
Here a , b and x are vectors and alpha is a scalar
I need a solution for x
• Aug 8th 2013, 01:30 PM
BobP
Re: Vectors
Take the scalar multiple of your equation, (call it eq(1)), with \$\displaystyle \underline{a},\$ and note that the triple scalar product is zero. Call this eq(2).
Take the vector multiple of eq(1) with \$\displaystyle \underline{a},\$ expand the triple vector product and substitute using eq(2). Call this eq(3).
Eliminate the vector product between eq's (1) and (3), and solve for \$\displaystyle \underline{x}.\$
• Aug 9th 2013, 05:54 PM
Kristen111111111111111111
Re: Vectors
Thank you so much for ypur help I got the answer :)
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Calculating Genetic Distances from RAPD Data
The image above shows DNA profiles of the population of var. arizonicus plants from the type locality (plants 4A-6D) for Primer G11. You may also choose to look at the image "Primer G11 4A-14B" to get a better look.
1) Score the DNA profiles for the presence or absence of the band. Only those bands which can be scored as clearly present or absent in each plant should be scored. For example, the 1.74 kb band is only present in plants 4D, 5B, and 5C. Score a "1" for presence, a "0" for absence. Do this for all scorable bands, as below.
Band (kb) 4-A 4-B 4-C 4-D 5-A 5-B 5-C 5-D 6-A 6-B 6-C 6-D 1.74 0 0 0 1 0 1 1 0 0 0 0 0 1.56 1 1 1 1 1 1 1 1 1 1 1 1 1.21 1 1 1 1 1 1 1 1 1 1 1 1 0.79 0 0 1 1 0 1 0 1 1 1 1 1 0.59 0 0 0 0 0 0 1 1 0 0 0 0 0.53 0 0 0 0 0 0 1 0 0 0 0 0 0.51 0 1 1 1 1 0 0 0 1 1 0 0
NOTE: Band 0.59 may look as though it is faintly present in plants 4A - 5B and 6A - 6D. However, these bands are slightly higher and probably derived from a different locus on the genome.
2) When you have scored all of your profiles for all plants, you can then begin your genetic analysis. Start by doing a pair-wise comparison, i.e. compare each plant to all others. For example, to start, compare plant #4A to plant #4B to determine the number of bands shared. For this primer, plants 4A and 4B share two bands out of a total of seven bands scored. Only count the shared presence of a band, and not the shared absence of a band. Next, compare 4A to 4C, and so on, until you have compared each plant to all of the others. Making a table with all of the plants across the top and the plants down the side will help you keep track of which plants you have compared.
Pair-wise Comparisons Table: Number of Bands Shared Out of 7 Bands
Band 4-A 4-B 4-C 4-D 5-A 5-B 5-C 5-D 6-A 6-B 6-C 6-D 4-A ----- 2 2 2 ____ ____ ____ ____ ____ ____ ____ ____ 4-B ----- ----- ____ ____ ____ ____ ____ ____ ____ ____ ____ ____ 4-C ----- ----- ----- ____ ____ ____ ____ ____ ____ ____ ____ ____ 4-D ----- ----- ----- ----- ____ ____ ____ ____ ____ ____ ____ ____ 5-A ----- ----- ----- ----- ----- ____ ____ ____ ____ ____ ____ ____ 5-B ----- ----- ----- ----- ----- ----- ____ ____ ____ ____ ____ ____ 5-C ----- ----- ----- ----- ----- ----- ----- ____ ____ ____ ____ ____ 5-D ----- ----- ----- ----- ----- ----- ----- ----- ____ ____ ____ ____ 6-A ----- ----- ----- ----- ----- ----- ----- ----- ----- ____ ____ ____ 6-B ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ____ ____ 6-C ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ____ 6-D ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- -----
3. To calculate the genetic distance of each band to all others you can use the simple equation:
For our example above, the genetic distance between plant #4A ("x") and plant #4B ("y") is:
For this very small number of bands scored, the calculation of genetic distance says that these plants are 33% different or 67% the same genetically. Obviously, the more bands you score and the more plants you have will increase the statistical significance of your calculations. You might use these calculations as a way to introduce the idea of statistical significance. For my research purposes, I will score about 100 bands to obtain statistically significant results.
4) Calculate the average genetic distance between plants within the same population. Use the chart below to determine to which population each plant belongs. Each number represents a different location, while each letter represents a different plant collected from that location. For example, four arizonicus plants (A, B, C, D) were collected from Location #4; they are named 4A, 4B, 4C and 4D.
Population Plant Numbers arizonicus 4, 5, 6 neomexicanus 13, 14, 15 triglochidiatis 7, 9, 10
Compare the average genetic distance within a population to the average genetic distance between populations. Is there more variation within the population or between different populations?
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https://www.instructables.com/Growing-Crystals/
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## Introduction: Growing Crystals!
In this Instructable, we will learn about super saturated solutions, and how to create crystals from them.
(Pictured above is an Alum crystal made using the explained method)
## Step 1: Understanding the Science
What happens when we dissolve salt in water? We add a few spoons, keep stirring till all the salt disappears. Salt, here, is the solute and water, the solvent. Together, they form a solution. There is a point when a solution contains more solute than the solvent can dissolve. This solution is known as a supersaturated solution. You normally cannot see the solute in a solution because it is fully dissolved. You are able to see the solute in a supersaturated solution because all of the solute doesn't fully dissolve into the solvent.
Here is a nice graph that I found online, when I was learning more about solutions. We can clearly see that solubility of a particular salt is directly proportional to temperature.So now, we will use the concept of dependence solubility of a solution on temperature to create crystals!
## Step 2: Materials Required
You will need the following to start our experiment:
1. A pair of gloves (optional, but recommended when dealing with salts like Borax)
2. Any salt soluble in water. I have used Borax here. Alternative salts that could be used are Alum, Magnesium Sulphate etc,
3. A cylindrical container, preferably transparent, so it would be easier to check progress without disturbing the system
4. Food colouring! (optional)
5. Water and a vessel that can be used to boil water
6. A scale (just a stick would do), some thread and a pair of scissors.
7. A sieve (optional)
With all this ready, lets move on!
## Step 3: Getting the Solvent Ready
Firstly, measure 3/4 of the cylindrical vessel with water (here water is the solvent). Pour the contents into another vessel, one with which we can boil the water. Now, heat the vessel till boil.
Science here: In this step, we boil the solvent in order to increase the solubility of the salt in the solvent, i.e. to create a super saturated solution, which dissolves more salt than at just room temperature.
## Step 4: Adding the Solute
Before handling with corrosive / dangerous salts, it is advisable that you glove up!
With a spoon, contentiously add salt to the boiling water. Please note, after the water starts to boil, do not switch of the flame, but leave it in sim and continuously stir will adding the salt.
Science here: There will be a point where you cannot add any more salt and the solution cannot accept any more salt. At this stage, the solution has reached saturation at its elevated temperature, but is at super saturation with respect to room temperature!
After this, switch of the stove, and a SMALL pinch of food colouring of your choice. I haven't done so here, but used green food colour in making those cool Alum salt crystals you saw in the thumbnail.
After doing so, pour the solution from the boiling mixture back into the cylindrical tumbler, through sieve, just to make sure no solid particle passes through.
## Step 5: Setting the Core
Once you are done with all the above steps, take a piece of thread of adequate length, and tie it up to the scale / stick as shown. Dip the thread into the solution with the scale supported by the vessel (as shown). It does not matter if the thread floats, but push it in as much as you can.
After doing so, leave the setup at an undisturbed place overnight. By undisturbed, I mean away from vibration.
Science here: The reason we do this is to provide a core foe the crystals to form around; similar to what dust particles do in clouds.
## Step 6: Fruits of Labour
After leaving the setup undisturbed for a while, it can be noted that a crystal has grown around the thread! Magical! Now remove the thread from the scale, cut off the excess and create your own crystal garden!
Science here: The reason we let the super saturated solution cool is, because while doing so, the temperature decreases and in turn, reduces the solubility of the solution. So the salt that would dissolve in the solute at a higher temperature, now have nowhere to go, as solubility decreases, so the crystallize out of the solution around the thread (like clouds). This is a gradual and slow process, that is why it is not advisable to store the solution in a fridge to cool, but best left at room temperature. Also, if there are any vibrations / shocks, it would disturb the crystal formation. So, no vibration means better and larger crystals!
You are now endowed with the secret knowledge and science of making super cool crystal! So what are you waiting for? Go ahead and start making 'em!
Arty tip: Use red colour food colouring to make red crystals, and instead of thread, use tin bent in the shape of a heart! You now have a red crystal shaped heart! ( I made this for my mom on mothers day and she loved it :-) )
Participated in the
Explore Science Contest 2017
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Decision Trees
# Decision Trees #
In this chapter we will treat a non-parametric method, the Decision Tree (DT) that is one of the most popular ML algorithms. They are used usually as components of ensemble methods. They are non-parametric models because they don’t need a predetermined set of parameters before training can start as in parametric models - rather the tree fits the data very closely and often overfits using as many parameters are required during training.
Its popularity stems also from yet another attribute that is becoming very important in the application of ML/AI in mission critical industries such as health: its ability to offer interpretable predictions to some degree that can also be introspected easily by humans. DTs can be used for both regression and classification - here we look at a classification use case but the principles are the same.
For example, let’s say that you had a basket of fruit in front of you, and you were trying to teach someone who had never seen these types of fruit before how to tell them apart. How could you do it? The answer is shown pictorially below.
A decision tree is a tree where each node represents a feature(attribute), each link (branch) represents a decision (rule) and each leaf represents an outcome(categorical for classification or continual for regression).
If your decision tree is good, you can now pick up an unknown piece of fruit and follow the flow chart to classify it. If your decision tree is bad, you can go down the wrong path and put something in the wrong category. For instance, if you didn’t know yellow apples existed when you built your decision tree, you might have assumed that all yellow fruit are either bananas or plantains.
In what follows, we focus on a dataset with $m=88$ and $4$ labels: Apples, Oranges, Bananas, Grapefruit. Each example has multiple features $n$: color, width and length.
Fruit Colors
Apples Red, Green, or Yellow
Oranges Orange
Bananas Yellow or Green
Grapefruit Orange or Yellow
Fruits dataset
If we are to draw separation lines on feature space of length ($x_1$) and width ($x_2$) without using an ML algorithm but by hand, we probably would come up with the picture below.
Draw by hand partition
Now, lets try to solve the same problem using an algorithm bearing in mind that many real-life data sets might have dozens or hundreds of different features.
## CART algorithm #
One of the most popular algorithms that implement decision trees is the Classification and Regression Tree (CART) algorithm.
NOTE: scikit-learn uses an optimised version of the CART algorithm; however, scikit-learn implementation does not support categorical variables for now.
At its heart, the algorithm implements a recursive binary partitioning of the input feature space. The feature space in the example above is $\mathbf{x} = (x_1, x_2, x_3)^T$ denoting length, width and color. Given a training dataset as usual $D={(\mathbf x_i, y_i)}$ for $i={1, \dots m}$, we need to come up with a close to optimal partitioning that minimizes the generalization (test) error.
We start at a root node that corresponds to the whole feature space (no partition) and design a test that is in its simplest form a conditional statement against a feature (a comparison if you like). Depending on the binary outcome of the test (either the input examples will satisfy the condition or not) we produce the corresponding child nodes each inheriting a subset of the input population and we repeat the exercise. The recursion stops when we reach the so called leaf nodes e.g. when the remaining examples in these nodes cannot be split further. We will come back at this terminal / leaf nodes later. An example tree and corresponding partition is shown in the two figures below.
Example tree - $x_1$ and $x_2$ based splits shown.
Partition for the example tree above.
The test specification consists of the variables $\theta_i$ that are are also called thresholds as well as the specific feature $x_k$ that is being selected for the test.
Lets see the three recursions of the algorithm as shown below.
First split
Second split
Third split
Final decision tree for the fruit classification problem
This brings up the question of how we select the test spec parameters $x_k$ and $\theta_k$ to minimize a certain metric that is dependent on the type of the problem we deal with - classification or regression.
## Selecting the feature $x_k$ to split #
To gauge which feature we will choose split requires a review of certain probabilistic concepts namely the concept of entropy. We can develop on top of entropy the concept of information gain that is pictorially explained using the example shown below
Information gain for two possible splits
The input dataset in this example has uniform distribution over classes - we have exactly the same number of points in each class. If we split the data horizontally (select feature $x_1$) this produces two sets of data. Each set is associated with a lower entropy (higher information, peakier class histograms) that is defined as usual
$$H(D) = − \sum_{c \in C} p( c ) \log(p( c ))$$
The entropy drops after any reasonable split as we exclude labels from the original set and end up with more homogenous sets of labels. The gain of information achieved by splitting the data into two parts (1 / 2 or left / right) is computed as
$$IG = H(D) − L(x_k, \theta_k)$$
$$= H(D) - \sum_{j \in {1,2}} \frac{|D^j|}{|D|} H(D^j)$$
where $|.|$ is the cardinality operator i.e. the number of elements in the corresponding set and $L$ is the loss function that the algorithm is searching for its minima by trying $(x_k, \theta_k)$ pairs. Apart from entropy we can also use another measure of impurity of the labels of the corresponding set - the Gini impurity measure that is also the default setting in scikit learn. Try to focus on understanding the entropy based measure as it is used in other algorithms that we cover and is completely analogous to Gini.
More often that not, the tree may grow to a point that we have overfitting. In DT implementations there are many hyperparameters that can control overfitting the DT to the data. Regularization therefore happens by (a) constraining the maximum growing depth, (b) limiting the splits to only the cases that we have a appreciable number of data points to split, (c) putting a cap in the number of leaf nodes (d) putting a cap in the number of data points per leaf node.
## Inference #
Training will result into the heuristically optimal decision tree. Inference is then straightforward as the input data will transverse the tree and find itself into a leaf node. We can also estimate the probability that an instance belongs to a particular class k or $p(c_k|\mathbf x)$. First it traverses the tree to find the leaf node for this instance, and then it returns the ratio of training instances of class k in this node. Therefore DTs can give us some form of confidence to the classification operation they do.
## Example #
A DT applied in the area of cybersecurity.
## References #
Some of the material used in this chapter is from (a) “ML with Random Forests and Decision Trees” by Scott Hartshorn, (b) Decision Forests for Classification, Regression, Density Estimation, Manifold Learning and Semi-Supervised Learning by Criminisi et.a.l.
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https://penzionzamecek.com/index.php/2022/06/13/the-basics-of-stud-poker/
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# The Basics of Stud Poker
A game of poker consists of five card combinations, called poker hands. The value of these hands inversely proportional to the mathematical frequency. Players make bets based on the hand they have and hope that the other players will match it. However, a player may also bluff by betting they have the best hand, in which case the other players must match the bet in order to win. For this reason, poker is a popular game for people to learn.
## The betting phase of poker
In the betting phase of poker, players place forced bets, called blind bets or ante bets, and raise their bets after each other. The winning hand is the one in which all players have not folded yet. The winner shares their winnings with the remaining players, and the process continues clockwise around the table. During the betting phase, a player may develop one or more poker hands. In general, betting intervals are a minimum of thirty seconds and extend for as long as a minute.
## Variations of poker
There are various variations of poker. Among them is stud poker. In this type of poker game, players must make the best five-card hand. The player with the highest hand wins the pot. This variation of the poker game is comparatively easier to understand and is also fun to play. Here are the main differences between stud poker and other variations. If you enjoy playing poker, you must try this variant! Let us look at some of them to get an idea of the game’s basic principles.
## Limits of bets
Poker games have betting limits, which govern the amount a player can open, raise, or call. The limit for each player varies, but generally it’s one bet per hand. Poker limits help players bet smarter and increase their chances of winning by preventing them from overbetting. Limits are important to learn, because a small number of players might end up winning big if they’re all playing at the same limit.
## Limits of raises in poker
While determining the limits of raises in poker, it’s important to note that they vary from game to game. In most games, the minimum bet is the big blind and players must raise at the same rate. In modern games, however, it is not uncommon for players to raise only if they have a higher poker hand. This means that, for example, a player who has a five-card stack must raise seven times as much as the big blind, or else it’s considered a check. If the opponent matches, then the raise will be counted as a re-raise, and the player must make a bet of the same amount.
## Identifying conservative players from aggressive players
When playing the game of poker, identifying conservative players from aggressive players is vital to winning. This can be done by looking at their physical appearance, including their neatly pressed shirt and tidy hair. Conservative players usually buy in quietly and get right to work once they are seated. By contrast, aggressive players tend to make huge bets early in the game. In addition to appearance, it’s important to remember that conservative players are typically more successful when they’re strong.
## Tells of a poker player
When a poker player gets distracted, it may be a sign of a weak hand or semi-bluffing. They may not be chatting as much as you would expect them to. Experienced players play more than one table, not just one. Changing chattiness between hands is a sign of a weak hand or semi-bluffing. Learn to recognize tells of a poker player so you can win the game.
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Game Development Reference
In-Depth Information
Theorem 1.19 (IEWDMS)
Suppose that G is a finite strategic game.
(i) If G is an outcome of IEWDMS from G and m is a Nash equilibrium
of G , then m is a Nash equilibrium of G.
(ii) If G is solved by IEWDMS, then the resulting joint strategy is a Nash
equilibrium of G.
Here is a simple application of this theorem.
Corollary 1.20 Every mixed extension of a finite strategic game has a
Nash equilibrium such that no strategy used in it is weakly dominated by a
mixed strategy.
Proof It su ces to apply Nash's Theorem 1.14 to an outcome of IEWDMS
and use item ( i ) of the above theorem.
Finally, observe that the outcome of IEWMDS does not need to be unique.
In fact, Example 1.9 applies here, as well.
1.5.3 Rationalizability
Finally, we consider iterated elimination of strategies that are never best
responses to a joint mixed strategy of the opponents. Following Bernheim
[1984] and Pearce [1984], strategies that survive such an elimination process
are called rationalizable strategies. 3
Formally, we define rationalizable strategies as follows. Consider a restric-
tion R of a finite strategic game G . Let
RAT ( R ):=( S 1 ,...,S n ) ,
where for all i
}
S i := {s i ∈ R i |∃m −i ∈× j = i Δ R j s i is a best response to m −i in G}.
Note the use of G instead of R in the definition of S i . We shall comment on
it below.
Consider now the outcome G RAT
∈{
1 ,...,n
starting with G .We
call then the strategies present in the restriction G RAT rationalizable .
We have the following counterpart of the IESDMS Theorem 1.16, due to
Bernheim [1984].
of iterating
RAT
Theorem 1.21
Assume a finite strategic game G.
3 More precisely, in each of these papers a different definition is used; see Apt [2007] for an
analysis of the conditions for which these definitions coincide.
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# Second derivative test
In geodesy and geophysics, the Bouguer anomaly (named after Pierre Bouguer) is a gravity anomaly, corrected for the height at which it is measured and the attraction of terrain. The height correction alone gives a free-air gravity anomaly.
## Anomaly
The Bouguer anomaly is related to the observed gravity ${\displaystyle g_{obs}}$ as follows:
${\displaystyle g_{B}=g_{obs}-g_{\lambda }+\delta g_{F}-\delta g_{B}}$
${\displaystyle g_{B}=g_{F}-\delta g_{B}}$
Here, ${\displaystyle g_{B}}$ is the Bouguer anomaly, ${\displaystyle g_{F}}$ is the free-air gravity anomaly, ${\displaystyle g_{\lambda }}$ is the correction for latitude (because the Earth is not a perfect sphere) and ${\displaystyle \delta g_{F}}$ is the free-air correction and ${\displaystyle \delta g_{B}}$ is the correction for terrain called the Bouguer reduction.
A Bouguer reduction is called simple or incomplete if the terrain is approximated by an infinite flat plate called the Bouguer plate. refined or complete Bouguer reduction removes the effects of terrain precisely. The difference between the two, the differential gravitational effect of the unevenness of the terrain, is called the terrain effect. It is always negative.[1]
## Simple reduction
The gravitational acceleration ${\displaystyle g}$ outside a Bouguer plate is perpendicular to the plate and towards it, with magnitude 2πG times the mass per unit area, where ${\displaystyle G}$ is the gravitational constant. It is independent of the distance to the plate (as can be proven most simply with Gauss's law for gravity, but can also be proven directly with Newton's law of gravity). The value of ${\displaystyle G}$ is 6.67 × 10−11 N m² kg−2, so ${\displaystyle g}$ is 4.191 × 10−10 N m² kg−2 times the mass per unit area. Using 1 Gal = 0.01 m/s² we get 4.191 × 10−5 mGal m² kg−1 times the mass per unit area. For mean rock density (2.67 g/cm³) this gives 0.1119 mGal/m.
The Bouguer reduction for a Bouguer plate of thickness ${\displaystyle \scriptstyle H}$ is
${\displaystyle \delta g_{B}=2\pi \rho GH,}$
where ${\displaystyle \rho }$ is the density of the material and ${\displaystyle G}$ is the constant of gravitation.[1] On Earth the effect on gravity of elevation is 0.3086 mGal/m decrease when going up, minus the gravity of the Bouguer plate, giving the Bouguer gradient of 0.1967 mGal/m.
More generally, for a mass distribution with the density depending on one Cartesian coordinate z only, gravity for any z is 2πG times the difference in mass per unit area on either side of this z value. A combination of two equal parallel infinite plates does not produce any gravity inside.
Organisational Psychologist Alfonzo Lester from Timmins, enjoys pinochle, property developers in new launch singapore property and textiles. Gets motivation through travel and just spent 7 days at Alejandro de Humboldt National Park.
42 year-old Environmental Consultant Merle Eure from Hudson, really loves snowboarding, property developers in new launch ec singapore and cosplay. Maintains a trip blog and has lots to write about after visiting Chhatrapati Shivaji Terminus (formerly Victoria Terminus).
## Notes
43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.
## References
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
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# Inequality graph
• Mar 3rd 2009, 07:41 PM
mgutmans02
Inequality graph
-x + y > 3
please list 4 points on a graph for x & y
thanks!!!(Rofl)
• Mar 3rd 2009, 08:00 PM
meymathis
It's not clear what you asking for. Please state the full problem. Are you asking for 4 points in the region? Are you asking for 4 points on the graph of the line? Does X and Y have to be integer valued?
• Mar 3rd 2009, 08:03 PM
craigmain
It's a simple line, (I think)
Hi,
The equation is the area above a line.
\$\displaystyle y>=x+3\$ consists of all the points including and above \$\displaystyle y=x+3\$ which is a line with gradient 1, x intercept \$\displaystyle (-3,0)\$ and y intercept \$\displaystyle (0,3)\$
• Mar 3rd 2009, 08:14 PM
meymathis
Yes it is a region bounded by a simple line. I can name lots of points that have something to do with the line and the region. It is not clear what the question is asking.
• Mar 3rd 2009, 08:59 PM
Jameson
Quote:
Originally Posted by craigmain
Hi,
The equation is the area above a line.
\$\displaystyle y>=x+3\$ consists of all the points including and above \$\displaystyle y=x+3\$ which is a line with gradient 1, x intercept \$\displaystyle (-3,0)\$ and y intercept \$\displaystyle (0,3)\$
I think the term "gradient" is too advanced for pre-calculus, although it's certainly correct and might be understood. Slope seems to be more commonly because early algebra only teaches about lines with constant gradients. I'm not correcting you, I just want to point out to take in mind the level of the poster.
Quote:
Originally Posted by meymathis
Yes it is a region bounded by a simple line. I can name lots of points that have something to do with the line and the region. It is not clear what the question is asking.
Agreed. Badly written question. "on a graph" could mean so many things.
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SOCR EduMaterials AnalysesCommandLineFDR Correction
(Difference between revisions)
Revision as of 05:23, 26 April 2012 (view source)IvoDinov (Talk | contribs)← Older edit Current revision as of 14:25, 3 February 2016 (view source)IvoDinov (Talk | contribs) Line 2,120: Line 2,120: ===Supplementary information=== ===Supplementary information=== + * [http://wiki.socr.umich.edu/index.php/SMHS_CorrectionMultipleTesting#FDR_.28false_discovery_rate.29 FDR Correction protocol, SMHS EBook] * [http://en.wikipedia.org/wiki/False_discovery_rate False Discovery Rate (FDR)] * [http://en.wikipedia.org/wiki/False_discovery_rate False Discovery Rate (FDR)] * [http://lib.stat.cmu.edu/~genovese/talks/hannover1-04.pdf FDR Tutorial] * [http://lib.stat.cmu.edu/~genovese/talks/hannover1-04.pdf FDR Tutorial]
Analyses Command-Line - False Discovery Rate (FDR) Corection
This page includes the information on how to use the SOCR Analyses library for the purpose of computing the False Discovery Rate (FDR) correction for multiple testing in volumetric and shape-based analyses. Access is provided via shell-based command-line interface on local machines. More information about other SOCR Analyses command-line interfaces is available here.
Introduction
In addition to the graphical user interfaces, via a web-browser, all SOCR Analyses allow command-line shell execution on local systems.
In a nutshell, FDR is a statistical method used in multiple hypothesis testing to correct the probability values for the effect of multiple comparisons. FDR controls the expected proportion of incorrectly rejected null hypotheses (i.e., FDR controls the type I error, α). For example, if 1000 observations were experimentally predicted to be different, and a maximum FDR for these observations was q=0.10, then 100 of these observations would be expected to be false positives. The q-value is the FDR analogue of the p-value. The q-value of an individual hypothesis test is the minimum FDR at which the test may be called significant. One can directly estimate q-values or alternatively, fix a level at which to control the FDR. See the references below for more information on FDR.
Suppose we perform m simultaneous hypothesis tests with a common procedure (test). For any given procedure, we classify the results as shown below. Only these values are observed m, D, and N. The other ones are unknown. Let T=true, F=False, D=Discovery, and N=Nondiscovery. FDR provides a protocol that balances the competing demands of sensitivity and specificity of the test.
Ho Retained Ho Rejected Total
Ho True TN FD To
Ho False FN TD T1
Total N D m
Then, FDR guarantees that the expectation of the relative-false-discovery (falsely-rejecting Ho, when it's true) is $E \left ( {FD \over D}\right ) \leq \alpha$.
The SOCR analysis implementation of FDR is based on this article, see references below.
FDR Usage
java -ms500m -mx1000m -cp /ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_core.jar:/ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.volume.Test_FDR -input Input_PValues_File -output Output_PValuesCorrected_File [-type type] [-fdr_rate q] -number integer -byteorder string
• Options:
• -help: prints usage
• -v: verbose mode (will print all input/output data in standard output stream.
• -input [InputFileName]: specify a file-name for an ASCII or binary file containing the raw p-values.
• -output [OutputFileName]: specify a file-name for the corresponding output file that will contain the FDR-corrected p-values.
• -type [0,1]: Type=0 is for ASCII text input file type, Type=1 is for binary 4Byte=Float Input file.
• -mask [Mask-filename]: specify a mask-file (0 or 1 intensities) restricting the voxels where the p-values are FDR corrected (optional),
• Mask for shape/geometry files are of the same type as the text p-values file (e.g., if the p-values file contains 1,000 p-values, 1 per line, then the mask must contain 1,000 lines of 0's and 1's
• Masks for raw binary volumes must be binary 1-byte volumes of the exact same dimensions as the input volume of raw 4-byte floating p-values.
• Dependence: If this option is included, it must be preceded by "-type".
• Caution: Be careful with using masks to FDR correct p-values. P-values outside of the mask are ignored. The skeleton of the Masking and FDR-correction code is as follows:
if (mask_volume.readUnsignedByte()>0) maskVolumeBoolean[i]=true; // get mask
if (maskVolumeBoolean[i]) maskedPValues[counter] = p_values[i]; // mask raw_p_values
thresholdedPMap = FDR(maskedPValues, fdr_rate).getThresholdedArray(); // FDR threshold masked p_values
if (maskVolumeBoolean[i]) FDR_corrected_PValues[i] = thresholdedPMap[counter]; // Save thresholded p_values
else FDR_corrected_PValues[i] = 1.0;
• -fdr_rate [double]: the False Discovery Rate level (e.g., 0.05)
• -number [int]: enter the number of P-values stored in the input file (output file will have the same number of floats).
• -byteorder string: string is one of {big, little, other}.
• big = BIG_ENDIAN processor
• little = LITTLE_ENDIAN processor
• other = default processor (java.nio.ByteOrder.nativeOrder())
• -verbose: to report all input and corrected-output p-values (avoid!)
Try-It-Online
You can test the FDR functionality using the Pipeline PWS Web-start server.
Shape-based p-Value FDR correction example
Edit a new file (FDR_Correction.csh) using any editor and paste this inside (make sure the file has executable permissions). Some operating systems/platforms may require variants of this (C-shell) script.
Results of FDR thresholding (right) the raw p-values (left) of comparing the local shape measures for hippocampal models for 2 groups
#!/bin/csh
date
java -ms500m -mx1000m -cp /ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_core.jar:/ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.volume.Test_FDR -input /ifs/ccb/CCB_SW_Tools/Statistics/SOCR_Statistics/SOCR_CSV_test_Scripts_Data/FDR_Test_Input_PValues.txt -output /ifs/ccb/CCB_SW_Tools/Statistics/SOCR_Statistics/SOCR_CSV_test_Scripts_Data/FDR_Test_PValuesCorrected_Output.txt -type 0 -mask /ifs/ccb/CCB_SW_Tools/Statistics/SOCR_Statistics/SOCR_CSV_test_Scripts_Data/FDR_Test_Input_PValues_MASK.txt -fdr_rate 0.05 -number 1000 -byteorder little
date
exit
Volume-based p-Value FDR correction example
Edit a new file (FDR_Correction.csh) using any editor and paste this inside (make sure the file has executable permissions). Some operating systems/platforms may require variants of this (C-shell) script.
Results of volumetric FDR thresholding (right) the raw p-values (left) obtained by comparing the Jacobians of the displacement fields of registering 2 groups of 3D MRI scans in a common anatomical atlas-space.
#!/bin/csh
date
java -ms500m -mx1000m -cp /ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_core.jar:/ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.volume.Test_FDR -input P_Value_mask_New_Reg_GROUP.img -output P_Value_mask_New_Reg_GROUP_FDRcorrected.img -type 1 -fdr_rate 0.05 -number 10648000 -byteorder little -mask P_Value_mask_New_Reg_GROUP_MASK.img
date
exit
Example
The table below shows the results of applying FDR-correction to the P_values obtained by non-parametric statistical test comparing the hippocampi surfaces (consisting of 1,00 vertices) of 2 groups of subject(AD and MCI). The FDR-corrected p-values were obtained by the command line included above.
Index Raw P-Values FDR-Corrected P-values
1.00000000 0.00041800 0.00041800
2.00000000 0.00060500 0.00060500
3.00000000 0.06424300 0.00000000
4.00000000 0.14547100 0.00000000
5.00000000 0.78607900 0.00000000
6.00000000 0.08867100 0.00000000
7.00000000 0.38782900 0.00000000
8.00000000 0.04057200 0.00000000
9.00000000 0.07980900 0.00000000
10.00000000 0.00072400 0.00072400
11.00000000 0.53734700 0.00000000
12.00000000 0.38782900 0.00000000
13.00000000 0.00060500 0.00060500
14.00000000 0.00715900 0.00000000
15.00000000 0.33592200 0.00000000
16.00000000 0.78607900 0.00000000
17.00000000 0.20827100 0.00000000
18.00000000 0.33592200 0.00000000
19.00000000 0.00530200 0.00000000
20.00000000 0.00829000 0.00000000
21.00000000 0.01104300 0.00000000
22.00000000 0.01458000 0.00000000
23.00000000 0.01270300 0.00000000
24.00000000 0.94099500 0.00000000
25.00000000 0.07980900 0.00000000
26.00000000 0.00013100 0.00013100
27.00000000 0.94099500 0.00000000
28.00000000 0.05127400 0.00000000
29.00000000 0.00122800 0.00000000
30.00000000 0.19098200 0.00000000
31.00000000 0.28871400 0.00000000
32.00000000 0.00060500 0.00060500
33.00000000 0.07980900 0.00000000
34.00000000 0.00389200 0.00000000
35.00000000 0.00086600 0.00086600
36.00000000 0.36129100 0.00000000
37.00000000 0.00023700 0.00023700
38.00000000 0.03182800 0.00000000
39.00000000 0.28871400 0.00000000
40.00000000 0.36129100 0.00000000
41.00000000 0.12008600 0.00000000
42.00000000 0.03597400 0.00000000
43.00000000 0.14547100 0.00000000
44.00000000 0.00019500 0.00019500
45.00000000 0.00007100 0.00007100
46.00000000 0.14547100 0.00000000
47.00000000 0.00389200 0.00000000
48.00000000 0.00715900 0.00000000
49.00000000 0.44435000 0.00000000
50.00000000 0.00389200 0.00000000
51.00000000 0.22666800 0.00000000
52.00000000 0.00530200 0.00000000
53.00000000 0.00829000 0.00000000
54.00000000 0.00240700 0.00000000
55.00000000 0.78607900 0.00000000
56.00000000 0.00530200 0.00000000
57.00000000 0.19098200 0.00000000
58.00000000 0.86287800 0.00000000
59.00000000 0.07980900 0.00000000
60.00000000 0.05127400 0.00000000
61.00000000 0.03597400 0.00000000
62.00000000 0.63921900 0.00000000
63.00000000 0.26687300 0.00000000
64.00000000 0.78607900 0.00000000
65.00000000 0.67489200 0.00000000
66.00000000 0.36129100 0.00000000
67.00000000 0.74839900 0.00000000
68.00000000 0.09831000 0.00000000
69.00000000 0.00023700 0.00023700
70.00000000 0.00172700 0.00000000
71.00000000 0.31172900 0.00000000
72.00000000 0.02809800 0.00000000
73.00000000 0.01104300 0.00000000
74.00000000 0.00122800 0.00000000
75.00000000 0.17477300 0.00000000
76.00000000 0.17477300 0.00000000
77.00000000 0.00283200 0.00000000
78.00000000 0.00530200 0.00000000
79.00000000 0.00204100 0.00000000
80.00000000 0.36129100 0.00000000
81.00000000 0.01669800 0.00000000
82.00000000 0.94099500 0.00000000
83.00000000 0.53734700 0.00000000
84.00000000 0.17477300 0.00000000
85.00000000 0.00013100 0.00013100
86.00000000 0.00028700 0.00028700
87.00000000 0.00454800 0.00000000
88.00000000 0.00957900 0.00000000
89.00000000 0.13230700 0.00000000
90.00000000 0.02809800 0.00000000
91.00000000 0.00204100 0.00000000
92.00000000 0.20827100 0.00000000
93.00000000 0.05127400 0.00000000
94.00000000 0.74839900 0.00000000
95.00000000 0.41552200 0.00000000
96.00000000 0.03597400 0.00000000
97.00000000 0.07980900 0.00000000
98.00000000 0.00023700 0.00023700
99.00000000 0.00829000 0.00000000
100.00000000 0.36129100 0.00000000
101.00000000 0.41552200 0.00000000
102.00000000 0.98031600 0.00000000
103.00000000 0.12008600 0.00000000
104.00000000 0.14547100 0.00000000
105.00000000 0.08867100 0.00000000
106.00000000 0.20827100 0.00000000
107.00000000 0.00389200 0.00000000
108.00000000 0.00023700 0.00023700
109.00000000 0.00034700 0.00034700
110.00000000 0.98031600 0.00000000
111.00000000 0.00086600 0.00086600
112.00000000 0.01908100 0.00000000
113.00000000 0.14547100 0.00000000
114.00000000 0.00332400 0.00000000
115.00000000 0.00034700 0.00034700
116.00000000 0.00103200 0.00000000
117.00000000 0.19098200 0.00000000
118.00000000 0.10876800 0.00000000
119.00000000 0.05127400 0.00000000
120.00000000 0.00060500 0.00060500
121.00000000 0.00204100 0.00000000
122.00000000 0.60436300 0.00000000
123.00000000 0.07168000 0.00000000
124.00000000 0.00389200 0.00000000
125.00000000 0.07980900 0.00000000
126.00000000 0.01908100 0.00000000
127.00000000 0.24619500 0.00000000
128.00000000 0.00389200 0.00000000
129.00000000 0.00616800 0.00000000
130.00000000 0.00103200 0.00000000
131.00000000 0.00122800 0.00000000
132.00000000 0.08867100 0.00000000
133.00000000 0.24619500 0.00000000
134.00000000 0.15961400 0.00000000
135.00000000 0.00060500 0.00060500
136.00000000 0.28871400 0.00000000
137.00000000 0.02175600 0.00000000
138.00000000 0.02175600 0.00000000
139.00000000 0.05745500 0.00000000
140.00000000 0.31172900 0.00000000
141.00000000 0.00103200 0.00000000
142.00000000 0.10876800 0.00000000
143.00000000 0.14547100 0.00000000
144.00000000 0.71131000 0.00000000
145.00000000 0.02809800 0.00000000
146.00000000 0.00240700 0.00000000
147.00000000 0.03597400 0.00000000
148.00000000 0.44435000 0.00000000
149.00000000 0.00122800 0.00000000
150.00000000 0.19098200 0.00000000
151.00000000 0.00060500 0.00060500
152.00000000 0.00007100 0.00007100
153.00000000 0.63921900 0.00000000
154.00000000 0.47428600 0.00000000
155.00000000 0.78607900 0.00000000
156.00000000 0.50529900 0.00000000
157.00000000 0.00616800 0.00000000
158.00000000 0.13230700 0.00000000
159.00000000 0.00829000 0.00000000
160.00000000 0.08867100 0.00000000
161.00000000 0.94099500 0.00000000
162.00000000 0.98031600 0.00000000
163.00000000 0.36129100 0.00000000
164.00000000 0.00016000 0.00016000
165.00000000 0.00145800 0.00000000
166.00000000 0.00023700 0.00023700
167.00000000 0.38782900 0.00000000
168.00000000 0.98031600 0.00000000
169.00000000 0.20827100 0.00000000
170.00000000 0.28871400 0.00000000
171.00000000 0.01669800 0.00000000
172.00000000 0.02475100 0.00000000
173.00000000 0.10876800 0.00000000
174.00000000 0.00072400 0.00072400
175.00000000 0.33592200 0.00000000
176.00000000 0.74839900 0.00000000
177.00000000 0.38782900 0.00000000
178.00000000 0.36129100 0.00000000
179.00000000 0.01908100 0.00000000
180.00000000 0.74839900 0.00000000
181.00000000 0.02475100 0.00000000
182.00000000 0.01270300 0.00000000
183.00000000 0.00028700 0.00028700
184.00000000 0.63921900 0.00000000
185.00000000 0.00240700 0.00000000
186.00000000 0.00715900 0.00000000
187.00000000 0.24619500 0.00000000
188.00000000 0.00715900 0.00000000
189.00000000 0.00240700 0.00000000
190.00000000 0.00616800 0.00000000
191.00000000 0.00957900 0.00000000
192.00000000 0.57038600 0.00000000
193.00000000 0.00829000 0.00000000
194.00000000 0.14547100 0.00000000
195.00000000 0.00122800 0.00000000
196.00000000 0.41552200 0.00000000
197.00000000 0.00332400 0.00000000
198.00000000 0.00010700 0.00010700
199.00000000 0.12008600 0.00000000
200.00000000 0.07168000 0.00000000
201.00000000 0.19098200 0.00000000
202.00000000 0.04565900 0.00000000
203.00000000 0.94099500 0.00000000
204.00000000 0.12008600 0.00000000
205.00000000 0.02809800 0.00000000
206.00000000 0.00103200 0.00000000
207.00000000 0.20827100 0.00000000
208.00000000 0.31172900 0.00000000
209.00000000 0.74839900 0.00000000
210.00000000 0.20827100 0.00000000
211.00000000 0.36129100 0.00000000
212.00000000 0.57038600 0.00000000
213.00000000 0.17477300 0.00000000
214.00000000 0.82426800 0.00000000
215.00000000 0.14547100 0.00000000
216.00000000 0.00060500 0.00060500
217.00000000 0.02809800 0.00000000
218.00000000 0.00005800 0.00005800
219.00000000 0.05127400 0.00000000
220.00000000 0.82426800 0.00000000
221.00000000 0.01908100 0.00000000
222.00000000 0.00389200 0.00000000
223.00000000 0.01669800 0.00000000
224.00000000 0.00616800 0.00000000
225.00000000 0.63921900 0.00000000
226.00000000 0.01458000 0.00000000
227.00000000 0.07168000 0.00000000
228.00000000 0.94099500 0.00000000
229.00000000 0.74839900 0.00000000
230.00000000 0.17477300 0.00000000
231.00000000 0.00103200 0.00000000
232.00000000 0.00957900 0.00000000
233.00000000 0.24619500 0.00000000
234.00000000 0.44435000 0.00000000
235.00000000 0.00050300 0.00050300
236.00000000 0.50529900 0.00000000
237.00000000 0.00957900 0.00000000
238.00000000 0.60436300 0.00000000
239.00000000 0.14547100 0.00000000
240.00000000 0.00145800 0.00000000
241.00000000 0.02809800 0.00000000
242.00000000 0.36129100 0.00000000
243.00000000 0.63921900 0.00000000
244.00000000 0.19098200 0.00000000
245.00000000 0.04057200 0.00000000
246.00000000 0.00103200 0.00000000
247.00000000 0.00204100 0.00000000
248.00000000 0.00041800 0.00041800
249.00000000 0.00060500 0.00060500
250.00000000 0.71131000 0.00000000
251.00000000 0.00204100 0.00000000
252.00000000 0.05745500 0.00000000
253.00000000 0.00028700 0.00028700
254.00000000 0.01908100 0.00000000
255.00000000 0.86287800 0.00000000
256.00000000 0.04565900 0.00000000
257.00000000 0.00050300 0.00050300
258.00000000 0.63921900 0.00000000
259.00000000 0.00389200 0.00000000
260.00000000 0.50529900 0.00000000
261.00000000 0.07980900 0.00000000
262.00000000 0.90181700 0.00000000
263.00000000 0.01270300 0.00000000
264.00000000 0.01458000 0.00000000
265.00000000 0.98031600 0.00000000
266.00000000 0.02809800 0.00000000
267.00000000 0.20827100 0.00000000
268.00000000 0.00389200 0.00000000
269.00000000 0.38782900 0.00000000
270.00000000 0.08867100 0.00000000
271.00000000 0.15961400 0.00000000
272.00000000 0.00013100 0.00013100
273.00000000 0.01270300 0.00000000
274.00000000 0.00332400 0.00000000
275.00000000 0.00957900 0.00000000
276.00000000 0.05745500 0.00000000
277.00000000 0.60436300 0.00000000
278.00000000 0.98031600 0.00000000
279.00000000 0.63921900 0.00000000
280.00000000 0.71131000 0.00000000
281.00000000 0.01908100 0.00000000
282.00000000 0.01908100 0.00000000
283.00000000 0.08867100 0.00000000
284.00000000 0.00086600 0.00086600
285.00000000 0.06424300 0.00000000
286.00000000 0.00086600 0.00086600
287.00000000 0.00240700 0.00000000
288.00000000 0.02175600 0.00000000
289.00000000 0.00530200 0.00000000
290.00000000 0.28871400 0.00000000
291.00000000 0.07168000 0.00000000
292.00000000 0.03597400 0.00000000
293.00000000 0.00016000 0.00016000
294.00000000 0.04565900 0.00000000
295.00000000 0.38782900 0.00000000
296.00000000 0.00008700 0.00008700
297.00000000 0.20827100 0.00000000
298.00000000 0.00204100 0.00000000
299.00000000 0.00145800 0.00000000
300.00000000 0.00172700 0.00000000
301.00000000 0.26687300 0.00000000
302.00000000 0.02175600 0.00000000
303.00000000 0.00240700 0.00000000
304.00000000 0.00530200 0.00000000
305.00000000 0.01270300 0.00000000
306.00000000 0.01458000 0.00000000
307.00000000 0.44435000 0.00000000
308.00000000 0.08867100 0.00000000
309.00000000 0.33592200 0.00000000
310.00000000 0.00829000 0.00000000
311.00000000 0.00041800 0.00041800
312.00000000 0.03597400 0.00000000
313.00000000 0.04565900 0.00000000
314.00000000 0.00283200 0.00000000
315.00000000 0.06424300 0.00000000
316.00000000 0.01104300 0.00000000
317.00000000 0.28871400 0.00000000
318.00000000 0.15961400 0.00000000
319.00000000 0.00028700 0.00028700
320.00000000 0.00829000 0.00000000
321.00000000 0.28871400 0.00000000
322.00000000 0.20827100 0.00000000
323.00000000 0.57038600 0.00000000
324.00000000 1.00000000 0.00000000
325.00000000 0.86287800 0.00000000
326.00000000 0.19098200 0.00000000
327.00000000 0.10876800 0.00000000
328.00000000 0.00283200 0.00000000
329.00000000 0.02475100 0.00000000
330.00000000 0.03182800 0.00000000
331.00000000 0.05127400 0.00000000
332.00000000 0.22666800 0.00000000
333.00000000 0.10876800 0.00000000
334.00000000 0.02175600 0.00000000
335.00000000 0.15961400 0.00000000
336.00000000 0.01270300 0.00000000
337.00000000 0.01908100 0.00000000
338.00000000 0.00008700 0.00008700
339.00000000 0.00072400 0.00072400
340.00000000 0.38782900 0.00000000
341.00000000 0.02809800 0.00000000
342.00000000 0.82426800 0.00000000
343.00000000 0.00103200 0.00000000
344.00000000 0.00034700 0.00034700
345.00000000 0.00010700 0.00010700
346.00000000 0.00616800 0.00000000
347.00000000 0.08867100 0.00000000
348.00000000 0.01104300 0.00000000
349.00000000 0.17477300 0.00000000
350.00000000 0.98031600 0.00000000
351.00000000 0.20827100 0.00000000
352.00000000 0.53734700 0.00000000
353.00000000 0.53734700 0.00000000
354.00000000 0.00103200 0.00000000
355.00000000 0.00028700 0.00028700
356.00000000 0.10876800 0.00000000
357.00000000 0.00050300 0.00050300
358.00000000 0.24619500 0.00000000
359.00000000 0.00041800 0.00041800
360.00000000 0.60436300 0.00000000
361.00000000 0.00086600 0.00086600
362.00000000 0.26687300 0.00000000
363.00000000 0.00086600 0.00086600
364.00000000 0.33592200 0.00000000
365.00000000 0.15961400 0.00000000
366.00000000 0.00829000 0.00000000
367.00000000 0.00145800 0.00000000
368.00000000 0.04057200 0.00000000
369.00000000 0.00086600 0.00086600
370.00000000 0.22666800 0.00000000
371.00000000 0.00829000 0.00000000
372.00000000 0.00122800 0.00000000
373.00000000 0.01270300 0.00000000
374.00000000 0.00122800 0.00000000
375.00000000 0.01669800 0.00000000
376.00000000 0.53734700 0.00000000
377.00000000 0.02809800 0.00000000
378.00000000 0.90181700 0.00000000
379.00000000 0.98031600 0.00000000
380.00000000 0.00283200 0.00000000
381.00000000 0.33592200 0.00000000
382.00000000 0.33592200 0.00000000
383.00000000 0.00034700 0.00034700
384.00000000 0.00389200 0.00000000
385.00000000 0.50529900 0.00000000
386.00000000 0.00829000 0.00000000
387.00000000 0.00172700 0.00000000
388.00000000 0.00034700 0.00034700
389.00000000 0.13230700 0.00000000
390.00000000 0.50529900 0.00000000
391.00000000 0.04565900 0.00000000
392.00000000 0.00034700 0.00034700
393.00000000 0.00072400 0.00072400
394.00000000 0.00283200 0.00000000
395.00000000 0.01104300 0.00000000
396.00000000 0.00072400 0.00072400
397.00000000 0.07980900 0.00000000
398.00000000 0.15961400 0.00000000
399.00000000 0.10876800 0.00000000
400.00000000 0.00332400 0.00000000
401.00000000 0.07980900 0.00000000
402.00000000 0.00283200 0.00000000
403.00000000 0.00616800 0.00000000
404.00000000 0.05745500 0.00000000
405.00000000 0.05745500 0.00000000
406.00000000 0.38782900 0.00000000
407.00000000 0.38782900 0.00000000
408.00000000 0.90181700 0.00000000
409.00000000 0.24619500 0.00000000
410.00000000 0.00028700 0.00028700
411.00000000 0.00957900 0.00000000
412.00000000 0.47428600 0.00000000
413.00000000 0.44435000 0.00000000
414.00000000 0.24619500 0.00000000
415.00000000 0.00172700 0.00000000
416.00000000 0.00240700 0.00000000
417.00000000 0.00204100 0.00000000
418.00000000 0.00034700 0.00034700
419.00000000 0.00013100 0.00013100
420.00000000 0.31172900 0.00000000
421.00000000 0.00072400 0.00072400
422.00000000 0.00086600 0.00086600
423.00000000 0.00060500 0.00060500
424.00000000 0.10876800 0.00000000
425.00000000 0.57038600 0.00000000
426.00000000 0.17477300 0.00000000
427.00000000 0.01104300 0.00000000
428.00000000 0.05127400 0.00000000
429.00000000 0.00008700 0.00008700
430.00000000 0.00019500 0.00019500
431.00000000 0.63921900 0.00000000
432.00000000 0.12008600 0.00000000
433.00000000 0.10876800 0.00000000
434.00000000 0.50529900 0.00000000
435.00000000 0.01104300 0.00000000
436.00000000 0.02475100 0.00000000
437.00000000 0.02475100 0.00000000
438.00000000 0.00086600 0.00086600
439.00000000 0.67489200 0.00000000
440.00000000 0.19098200 0.00000000
441.00000000 0.38782900 0.00000000
442.00000000 0.20827100 0.00000000
443.00000000 0.57038600 0.00000000
444.00000000 0.07168000 0.00000000
445.00000000 0.01908100 0.00000000
446.00000000 0.26687300 0.00000000
447.00000000 0.00086600 0.00086600
448.00000000 0.01669800 0.00000000
449.00000000 0.01669800 0.00000000
450.00000000 0.04565900 0.00000000
451.00000000 0.41552200 0.00000000
452.00000000 0.00103200 0.00000000
453.00000000 0.12008600 0.00000000
454.00000000 0.01270300 0.00000000
455.00000000 0.00072400 0.00072400
456.00000000 0.00172700 0.00000000
457.00000000 0.19098200 0.00000000
458.00000000 0.03597400 0.00000000
459.00000000 0.38782900 0.00000000
460.00000000 0.00616800 0.00000000
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Validation
Pipeline FDR Validation Protocol.
The SOCR FDR calculator was validated using the following protocol. The following ZIP archive contains data, validation protocol and a pipeline graphical workflow implementation of this validation protocol. This protocol uses the following data:
• Input Files:
Raw_P.txt - 100 random numbers in (0,1)
• The protocol generates the following output FDR corrected p-value results:
FDR_q0.05_Raw_P.txt
• Explicit validation scripts:
Calculation of FDR corrected p-values without masking (output: FDR_q0.05_Raw_P.txt):
java -ms500m -mx1000m -cp /usr/local/loniJars/SOCR_Statistics/SOCR_core.jar:/usr/local/loniJars/SOCR_Statistics/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.volume.Test_FDR -input Raw_P.txt -output FDR_q0.05_Raw_P.txt -type 0 -fdr_rate 0.05 -number 100 -byteorder little
java -ms500m -mx1000m -cp /usr/local/loniJars/SOCR_Statistics/SOCR_core.jar:/usr/local/loniJars/SOCR_Statistics/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.volume.Test_FDR -input PreMasked_41_P.txt -output FDR_NoMask_PreMaskedRaw_P.txt -type 0 -fdr_rate 0.05 -number 52 -byteorder little
Calculation of FDR corrected p-values by providing the mask file to the FDR-calculator (output: FDR_UsingMask_Raw_P_values.txt):
java -ms500m -mx1000m -cp /usr/local/loniJars/SOCR_Statistics/SOCR_core.jar:/usr/local/loniJars/SOCR_Statistics/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.volume.Test_FDR -input Raw_P.txt -output FDR_UsingMask_Raw_P_values.txt -type 0 -fdr_rate 0.05 -number 100 -byteorder little -mask Mask.txt
| 20,405
| 44,130
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.5625
| 3
|
CC-MAIN-2022-05
|
latest
|
en
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# Posts by marla
Total # Posts: 36
algebra
Enter the ratio as a fraction in lowest terms 5 ft. to 26 inches
Algebra
V/6=3/7
science
what are the different daily observations in seed germination. What are the factors and its effect on germination? Data Chart: Cup #1 Seeds with water and sunlight Cup #2 Seeds with water and no sunlight Cup #3 Seeds with sunlight and no water OBSERVATION Cup #1 / Cup #2/ Cup...
Trigonometry
5. A backpacker notes that from a certain point on level ground, the angle of elevation to a point at the top of a tree is 34o. After walking 50 closer to the tree, the backpacker notes that the angle of elevation is 54o. Find the height of the tree.
Math
7 1/6
algebra
49x^2+21x
algebra
60x^2-15x
Science
1) the plants on the floor wouldn't get as much light, since the tree branches above them "hog" the light and cover the floor in shade. The floor leaves get less light, the canopy get more. 2) Sea grass in muddied water will have the same problem as 1). Their ...
medical transcription
Could someone review this and make sure it appears to be correct? Had a repeat magnetic resonance imaging (MRI) of the head. She had a lesion seen in the right frontal lobe on previous computed tomography (CT) and MRI scanning. This lesion has not changed at all over time. I ...
Assume a bank loan requires an interest payment of \$85 per yr and the principal pymt of \$1k at the end of the 8 yr life, How much could this loan be sold for to another bank of similar quality carried an 8.5% rate
chemistry
i need to prepare 25.00 mL of a solution of a certain vitamin at a concentration of 2.00 mg/ml. how do i find the weight of the vitamin needed in the solution
chemisty
I have data for titration of a strong base against a strong acid when I added volumes of NaOH 1 mL at a time to 50.00 mL of HCl to see the change in pH. One of the questions says to find the molarity of the HCl solution but how do I do this with the data.
psy 202
8. What is the general term Freud used to describe the process whereby the mind protects itself against distress? (Points : 1) Compensation Defense mechanism Rationalization Displacement
Algebra1
graph of two infinite lines that intersect at a point. One line is solid and goes through the points (0, 2) (negative 2, 4) and is shaded in below the line. The other line is solid, and goes through the points (0, 2) (2, 0) and is shaded in below the line. A. y less than or ...
Composition II
I have to write a causal analysis to the Frederick Douglass speech "What to the slaves is the Fourth of July?" I am having trouble doing this.
math
The floor tile is one square foot. The kitchen floor was covered with 100 floor tiles. What was the area of the kitchen floor in square inches?
math
I believe the answer was IF so not OF so, thank you.
Math
which of the following sums is larger and by how much? round your answer to 4 decimal places if necessary 1. 7 over Sigma 15/4(pi/10)^n 2. 11 over Sigma 17/4(4/13)^n please help!
help math
a. what is the general term of the following series? 60/121-30/11+15-... +219 615/16 b. how many terms does the series have
Christian Couseling
What are the major themes of The Discipleship Counseling Handbook by Neil T. Anderson
statistics
Container 1 has 8 items, 3 of which are defective. Container 2 has 5 items, 2 of which are defective. If one item is drawn independently from each container: Find the probability distribution for X defined as the number of defective items drawn(Hint: You have to use both ...
algebra
I am having trouble with the following : find the slope and the y intercept of the line 7x=6y+12. I have the slope as 3 and the y intercept (0,3)
PSYCHOLOGY
8. What is the general term Freud used to describe the process whereby the mind protects itself against distress? (Points : 1) Compensation Defense mechanism Rationalization Displacement
Chemistry
Does anybody know the principle quantum number of an atom that has s, p, d, f orbitals? I'm pretty sure the number is inbetween 2-4, but I am unsure of how to figure out the principle quantum number just using that information. Thanks in advance!
english
fruit from argentina
literary
Throughout chapter 7, what does the author do to help you learn about Stanley's "no good dirty rotten pig stealing great great grandfather
Math
Four hundred people apply for three jobs. 130 of the applicants are women. (a) If three persons are selected at random, what is the probability that all are women? (Round the answer to six decimal places.) (b) If three persons are selected at random, what is the probability ...
Math 101
Debra is buying prizes for a game at her school's fundraiser. The game has three levels of prizes, and she has already bought the second and third prizes. She wants the first prize to be nice enough to attract people to the game. The game's manufacturer has supplied ...
Accounting
Help with Crosby Corporation? Can you please help me. I am completly stuck. Prepare a statement of cash flows for the Crosby Corporation. Follow the general procedures indicated in Table 2–10 on page 38 . Statement of cash flows (L04) Current Assets Liabilities Cash...
Math
Steves mom age is 7 years less than 3 times steves age. The sum of their ages is 65 years. Find their ages.
english
australia
english
Everything is the same nothing has changed and still he continues to hate his trade and home.
english
I don't understand this quote from Great Expectations. Can you help me paraphase it? So unchanging was the dull old house,the yellow lights in the darkened room the faded spectre in the chair by the dressing table glass, that I felt as if the stopping of the clocks had ...
English
I don't understand this quote from Great Expectations. Can you help me paraphase it?
english
So unchanging was the dull old house,the yellow lights in the darkened room the faded spectre in the chair by the dressing table glass, that I felt as if the stopping of the clocks had stopped time in that mysterious place , and while I had everything else outside, it grew ...
Food Webs and Food Chains
does any one know any food chains or food webs? its an emergency! http://en.wikipedia.org/wiki/Food_chain
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https://gateoverflow.in/blog/tag/iit-interviews
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# Recent posts tagged iit-interviews
1
i think the language would be L= 00(0000)* if L= 00 + (0000)* then 0000 cant reaches the final state .
2
Yeah why not you have found a solution it cab be one of the option but always choose the best one
3
return res * x; where x is a pointer while multiplication on pinters is not valid so it is still ambigious
4
If you put "Null" as introducer it will work. But by self introduction I mean "john" introducing himself- introducer will be "john" for customer "john". But I guess this needn't be considered unless specified as by the meaning of "introduction" we can assume it is a different person.
7
A)The n^{th} statement in a list of 100 statements is: "Exactly n of the statements in this list are false." 1)Exactly 1 of the statements in this list are false. 2)Exactly 2 of the statements in this list are false. 3)Exactly 3 of the statements in this list are false. . . ... (100) will be false. B)Same way we can solve it. So we will get 1-50 are true and 51-100 are false. C)This cannot happen.
8
Barber is one out of set of affected people.. if barber is in trouble then above statement is illogical.. if he is in trouble then der is a barber who save him but that saver barber is in trouble. i.e. no such barber exists.
10
bhai integration is area only http://www.mathsisfun.com/calculus/integration-introduction.html
11
12
If $g(x) = 1 - x$ and $h(x) = \frac{x}{x-1}$, then $\frac{g(h(x))}{h(g(x))}$ is: $\frac{h(x)}{g(x)}$ $\frac{-1}{x}$ $\frac{g(x)}{h(x)}$ $\frac{x}{(1-x)^{2}}$
13
Is there any SQL command which belongs to both DDL and DML?
14
Answer is D. $L_1$ is context-free and hence recursive also. Recursive set being closed under complement, $L_1$' will be recursive. $L_1$' being recursive it is also recursively enumerable and Recursively Enumerable set is closed under Union. So, $L_1' \cup L_2$ is recursively enumerable. ... $L_2$')' $= L_2$ is also recursive which is not the case here. So, $II$ is also false.
15
$\lim_{x\rightarrow \infty } x^{ \tfrac{1}{x}}$ is $\infty$ 0 1 Not defined
16
2 is correct answer. fig 2, b and c have {f,g} as upperbound. but for the graph to be a lattice it should have a least upper bound. Since b and c have two upper bounds they cannot have a least upper bound<which is always unique for a pair for vertices>. In ... {f,g}. therefore it is not a join semilattice(every pair of element should have a least upper bound). henceforth it is also not a lattice
18
L2 is surely regular and I would choose L1 to be non-regular. Is the answer correct?
19
$(C012.25)_H - (10111001110.101)_B$ $= 1100\;0000\;0001\;0010.\;0010\;0101$ $- 0000\;0101\;1100\;1110.\;1010\;0000$ $= 1011\;1010\;0100\;0011.\;1000\;0101$ $= 1\;011\;101\;001\;000\;011 .\;100\;001\;010$ $= (135103.412)_o$ Binary subtraction is like decimal subtraction:$0-0 = 0, 1-1 = 0, 1-0 = 1, 0-1 = 1$ with $1$ borrow. Correct Answer: $A$
20
SLR paper is more powerful than LALR . False . LALR parser is more powerful than Canonical LR parser . False . Canonical LR parser is more powerful than LALR parser. True. The parsers SLR, Canonical CR, and LALR have the same power. False. answer - C
21
How query 2 will not produce the correct result? I am getting the desired result from query 2.
22
Instead of finding a function 'f' for the first MUX (in terms of z,x and y) and the second MUX (in terms of f , x and y), we can logically find out the answer.We can put in 4 values of A and B and analyse the output C. We can get the values appropriate values of x and y by drawing and comparing with a 2x1 MUX when A and B are given as inputs respectively.
24
Later I got another idea to solve it A' + AB',apply distributive property, that comes to (A'+A)(A'+B') = A'+B'
25
Solution 48: spills to m/m while execution of program how many times m/m will be hit, we have 2 regiters Program Execution c = a + b; R2 <- R1 + R2 d = c * a; [m/m spill] <- R2*R1(we need R & R1 further value cant be replaced) e = c + a; [m/m spill] <- R2+R1 x = c * c; R2 < ... } else { d = d * d; R3<- R3*R3 e = e * e; R4<- R4*R4 } If 4 registers are used, m/m spills will be 0, hence answer is (B)
26
Time without pipeline $=6 \text{ stages}=6 \text{ cycles}$ Time with pipeline $=1+\text{stall freqency}\times \text{stall cycle}$ $=1+.25\times 2$ $=1.5$ Speed up $=\dfrac{6}{1.5}=4$
27
(c) Every finite subset of a regular set is regular this is false example:a^n b^n from regular set (a+b)* is not regular
28
Level order traversal of a rooted tree can be done by starting from the root and performing preorder traversal in-order traversal depth first search breadth first search
To see more, click for the full list of questions or popular tags.
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Over the past decade, passive income strategies have made a prominent name for themselves in the investing community. The name says it all: passive income. Once you have conducted the appropriate due diligence, you essentially just need to wait for the checks to start coming in. However, for one reason or another, there are those that are still skeptical of passive income strategies. Whether or not this is ignorance or trepidation, there is only one thing you can do to overcome any reservations you may have: education. The more you learn about passive income, the more inclined you may be to exercise one of its strategies.
If you have specialized knowledge in a certain topic, you can put together an online course to teach others. For example, if you have experience in real estate investing, you can create an online course “Real Estate Investing 101”. The benefit of an online course is that once you create the course material, you can sell it to as many people as you want.
4. Calculate how much passive income you need. It's important to have a passive-income goal — otherwise, it's very easy to lose motivation. A good goal is to try to generate enough passive income to cover basic living expenses such as food, shelter, transportation, and clothing. If your annual expense number is \$30,000, divide that figure by your expected rate of return to see how much capital you need to save. Unfortunately, you've got to then multiply the capital amount by 1.25 to 1.5 to account for taxes.
The members and brokers that Brad recruited, as well as the members and brokers that those people recruited, were considered Brad’s “downline.” At the time of the divorce, Brad’s downline consisted of thousands of members and brokers, earning Brad a residual income of about \$27,000 per month. The trial court was tasked with determining just how to divide the residual income, generated by Brad’s downline, between the two parties.
Residual income is calculated as net income less a charge for the cost of capital. The charge is known as the equity charge and is calculated as the value of equity capital multiplied by the cost of equity or the required rate of return on equity. Given the opportunity cost of equity, a company can have positive net income but negative residual income.
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https://usethinkscript.com/threads/define-sell-criteria-when-profit-target-is-reached.9581/
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# Define Sell Criteria When Profit Target Is Reached
#### AHARDGE
##### New member
Hello,
Can someone please assist me with defining the sell criteria for this code? I would like for the criteria to sell when the profit has reached 20%. I know the criteria that I have now is not correct because my code isn't executing the sell. Here is what I have so far:
def buy = MACDHistogramCrossover("crossing type" = "Negative to Positive") is true;
def sell = profitLossMode.COST_BASIS) ;
Solution
Hello,
Can someone please assist me with defining the sell criteria for this code? I would like for the criteria to sell when the profit has reached 20%. I know the criteria that I have now is not correct because my code isn't executing the sell. Here is what I have so far:
def buy = MACDHistogramCrossover("crossing type" = "Negative to Positive") is true;
def sell = profitLossMode.COST_BASIS) ;
One way to do it is to set an array that maintains the purchase price and compares the close (or high) to that...
Code:
``````def buy = MACDHistogramCrossover("crossing type" = "Negative to Positive") is true;
def cost = if buy[1] then open else cost[1];
def sell = if high > cost * 1.2 then high else double.nan;``````
Hello,
Can someone please assist me with defining the sell criteria for this code? I would like for the criteria to sell when the profit has reached 20%. I know the criteria that I have now is not correct because my code isn't executing the sell. Here is what I have so far:
def buy = MACDHistogramCrossover("crossing type" = "Negative to Positive") is true;
def sell = profitLossMode.COST_BASIS) ;
One way to do it is to set an array that maintains the purchase price and compares the close (or high) to that...
Code:
``````def buy = MACDHistogramCrossover("crossing type" = "Negative to Positive") is true;
def cost = if buy[1] then open else cost[1];
def sell = if high > cost * 1.2 then high else double.nan;``````
adjust to your needs. I chose the condition for cost to be if the previous bar says buy then pick this bar's open. ToS defaults to
it does essentially the same thing.
-mashume
87k+ Posts
335 Online
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https://mathematica.stackexchange.com/questions/184147/how-to-stop-derivative-evaluation
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# How to stop derivative evaluation?
It is posible to stop derivative evaulation in an expression?
Given
A = f[x]g[x];
F = W[t, n] /. W[x_, n_] -> D[A, {x, n}]
I would like F to be PartD(t, n)(f(t)g(t)) with the derivative not calculated, and when I evaluate
F /. n -> 0
I would like to get the general partial derivative
f[t]g[t]
P.S. Sorry for language mistakes.
• Could just do the usual Leibniz rule: Sum[Binomial[n, k] Derivative[n - k][f][t] Derivative[k][g][t], {k, 0, n}] – J. M.'s ennui Oct 18 '18 at 10:09
• A=f[x]*g[x] is simple example. For general case i dont know how A look like. – zbyneksmetana Oct 18 '18 at 10:14
• So you want something like Inactive[D][A, {x, n}]? – J. M.'s ennui Oct 18 '18 at 10:16
• Thanks ... maybe it will help. – zbyneksmetana Oct 18 '18 at 10:22
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https://www.teachoo.com/1293/930/Ex-12.1--6---An-isosceles-triangle-has-perimeter-30-cm/category/Finding-area-of-triangle/
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Finding area of triangle
Chapter 12 Class 9 Herons Formula
Concept wise
This video is only available for Teachoo black users
Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
### Transcript
Ex 12.1, 6 An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle. Area of triangle = (s(s a)(s b)(s c)) Here, s is the semi-perimeter, and a, b, c are the sides of the triangle Given triangle is isosceles In isosceles triangle, two sides are equal So, a = b = 12 cm and Perimeter = 30 cm Semi-Perimeter = s =Perimeter/2 s = 30/2 s = 15 cm We need to find c Perimeter = 30cm a + b + c = 30 cm 12 cm+ 12 cm + c = 30 cm 24 + c = 30 c = 30 24 c = 6 cm Area of triangle = ( ( )( )( )) Putting a = 12 cm, b =12 cm, c = 6 cm & s = 15 cm = (15(15 12)(15 12)(15 6)) = (15(3)(3)(9)) = (15 (9)(9)) = ("9" 9) 15 = ("92" ) 15 = (9) 15 = 9 15 Thus, Area = 9 15 cm2
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# Make Informed Financial Decisions with the MIRR Calculator
The MIRR calculator is a powerful financial analysis tool that helps individuals evaluate investment opportunities and make informed decisions. Unlike the traditional Internal Rate of Return (IRR), which does not consider the reinvestment of cash flows, the MIRR takes into account the rate at which cash inflows are reinvested. This provides a more realistic assessment of investment profitability.
By inputting the initial investment, cash inflows, cash outflows, and the discount rate for reinvestment and financing, the MIRR calculator calculates the MIRR. If the calculated MIRR is higher than the required rate of return, it indicates that the investment is expected to generate a profit. This makes the MIRR calculator an invaluable resource for assessing investment opportunities.
### Key Takeaways:
• The MIRR calculator evaluates investment opportunities by considering the reinvestment of cash flows.
• It provides a more realistic assessment of investment profitability compared to the traditional IRR.
• Input the initial investment, cash inflows, cash outflows, and the discount rate to calculate the MIRR.
• If the calculated MIRR is higher than the required rate of return, the investment is expected to generate a profit.
• The MIRR calculator is particularly useful for projects with complex cash flows and significant financing needs.
## Understanding the Modified Internal Rate of Return
The Modified Internal Rate of Return (MIRR) is a financial metric that takes into account the reinvestment of cash flows, providing a more realistic evaluation of investment opportunities. Unlike the traditional Internal Rate of Return (IRR), which assumes that cash inflows are reinvested at the same rate of return, the MIRR considers the actual rate at which cash inflows can be reinvested.
According to Investopedia, the MIRR formula is calculated by finding the rate of return that equates the present value of cash inflows to the present value of cash outflows, considering both the reinvestment and financing rates. This formula allows investors to account for the time value of money and the different rates at which cash flows can be reinvested, leading to a more accurate assessment of investment profitability.
When using the MIRR calculation, it can be helpful to utilize spreadsheet software such as Microsoft Excel. Excel provides built-in functions that simplify the MIRR calculation, allowing for easy analysis of investment opportunities. By inputting the initial investment, cash inflows, cash outflows, and the respective reinvestment and financing rates, Excel can quickly calculate the MIRR, saving time and effort. The table below illustrates an example of using Excel to calculate the MIRR:
Year Cash Flows
0 (Initial Investment)
1 \$10,000
2 \$15,000
3 \$5,000
(Reinvestment Rate)
(Financing Rate)
By using the MIRR calculator or spreadsheet software like Excel, investors can effectively evaluate investment opportunities by considering the reinvestment of cash flows. The MIRR calculation provides a more accurate assessment of investment profitability, helping users make informed financial decisions. When analyzing investment opportunities, it is important to consider other financial metrics such as net present value (NPV) and payback period to gain a comprehensive understanding. However, the MIRR remains a valuable tool in investment evaluation.
## How the MIRR Calculator Works
The MIRR calculator allows users to input key financial data, such as initial investment, cash inflows, and cash outflows, to calculate the Modified Internal Rate of Return. Unlike other financial metrics, the MIRR takes into account the reinvestment of cash flows at a separate rate of return, providing a more accurate assessment of an investment’s profitability.
To use the MIRR calculator, users simply need to input the relevant financial data for each period. This includes the initial investment, as well as the cash inflows and outflows for each period. Additionally, users must specify the discount rate for reinvestment and financing.
Once the data is inputted, the MIRR calculator performs the necessary calculations to determine the Modified Internal Rate of Return. If the calculated MIRR is higher than the required rate of return, it indicates that the investment is expected to generate a profit. This allows users to evaluate the potential profitability of an investment and make more informed financial decisions.
### Table: Example MIRR Calculation
Period Cash Inflows Cash Outflows
0 -100,000 0
1 30,000 10,000
2 40,000 15,000
3 50,000 0
Note: The above table represents an example calculation and does not represent real data.
Using the data from the table, the MIRR calculator determines the Modified Internal Rate of Return based on the cash inflows, outflows, and the specified discount rate. This allows users to evaluate the profitability of an investment and make informed financial decisions.
## Evaluating Investment Profitability with MIRR
By calculating the MIRR, investors can determine whether an investment is expected to be profitable based on the required rate of return. The Modified Internal Rate of Return (MIRR) takes into account the reinvestment of cash flows at a separate rate, providing a more accurate assessment of investment profitability compared to traditional metrics.
Using the MIRR calculator, investors can input the initial investment, cash inflows, and cash outflows for each period, along with the discount rate for reinvestment and financing. The calculator then calculates the MIRR, which can be compared to the required rate of return to evaluate the investment’s potential profitability.
Complex cash flows or projects that involve significant financing can benefit greatly from the MIRR calculation. The MIRR helps investors make informed decisions by considering the reinvestment of cash flows and providing a more realistic assessment of the investment’s value and expected return. With the MIRR calculator, investors can easily assess the potential profitability of investments and make smarter financial decisions.
Investment Details Cash Flows Discount Rate MIRR
Initial Investment Cash Inflows Cash Outflows Required Rate of Return
\$10,000 \$2,000, \$3,000, \$4,000 \$1,000, \$1,500, \$2,000 10%
Consider the example above, where an initial investment of \$10,000 generates cash inflows of \$2,000, \$3,000, and \$4,000 in subsequent periods, while incurring cash outflows of \$1,000, \$1,500, and \$2,000. By inputting these values into the MIRR calculator, along with the required rate of return of 10%, investors can determine whether the investment is expected to be profitable based on the calculated MIRR.
## Advantages of Using the MIRR Calculation
The MIRR calculation offers several advantages in financial analysis and investment evaluation, providing a more accurate assessment of investment value. Unlike the traditional Internal Rate of Return (IRR), which assumes reinvestment of cash flows at the same rate, the MIRR considers the reinvestment of cash inflows at a separate rate of return. This is particularly beneficial when evaluating investment opportunities with different rates of return.
By incorporating the reinvestment of cash flows, the MIRR takes into account the time value of money and provides a more realistic measure of profitability. It enables investors to make informed decisions by considering the potential returns on reinvested cash flows. This helps in accurately evaluating the actual value and potential return of an investment, ensuring better decision-making.
Another advantage of using the MIRR calculation is its suitability for complex cash flows and projects that involve substantial financing. Unlike the traditional IRR, which may not accurately reflect the profitability of such investments, the MIRR considers the unique cash flow patterns and financing requirements. This makes it an ideal tool for evaluating projects with multiple cash inflows and outflows.
Furthermore, the MIRR calculation can be easily performed using spreadsheet software, adding to its convenience and accessibility. With online MIRR calculators available, users can simply input the required data, including the initial investment, cash inflows, cash outflows, and the discount rate for reinvestment and financing. The calculator then provides the MIRR, allowing users to evaluate investment profitability quickly and efficiently.
Advantages of Using the MIRR Calculation
The MIRR provides a more accurate assessment of investment value by considering the reinvestment of cash inflows at a separate rate of return.
The MIRR incorporates the time value of money, enabling investors to make informed decisions based on potential returns on reinvested cash flows.
The MIRR is suitable for evaluating complex cash flows and projects that involve substantial financing, providing a more realistic measure of profitability.
The MIRR calculation can be easily performed using spreadsheet software or online MIRR calculators, offering convenience and accessibility.
## Complex Cash Flows and Financing Considerations
The MIRR calculation is particularly useful for evaluating investments with complex cash flows and projects that involve substantial financing. These types of investments often require a comprehensive analysis to accurately assess their potential profitability. By considering the reinvestment of cash flows at a separate rate of return, the MIRR provides a more accurate measure of the investment’s value and the expected return.
When dealing with complex cash flows, it can be challenging to determine the true profitability of an investment. The MIRR calculator simplifies this process by allowing users to input the various cash inflows and outflows for each period. By taking into account the timing and magnitude of these cash flows, the calculator calculates the MIRR, providing a clear indication of the investment’s profitability.
In addition, projects that require substantial financing can benefit greatly from MIRR calculations. The MIRR takes into consideration the cost of financing, allowing investors to evaluate the impact of interest rates and other financing costs on the investment’s potential return. By accurately accounting for these factors, the MIRR gives a more realistic assessment of the investment’s profitability, enabling investors to make more informed decisions.
Benefits of MIRR Calculation for Complex Cash Flows and Financing Considerations:
More accurate assessment of investment value
Clear indication of potential profitability
Ability to evaluate impact of financing costs
By using the MIRR calculator, investors can navigate the complexities of evaluating investments with complex cash flows and significant financing requirements. The calculator provides a reliable metric for assessing the investment’s profitability and potential return. Alongside other financial metrics, such as net present value (NPV) and payback period, the MIRR calculation offers a comprehensive evaluation that helps guide the decision-making process. It is an indispensable tool for investors looking to make informed choices and maximize their returns.
## Considering Other Financial Metrics
When evaluating investment opportunities, it’s crucial to consider other financial metrics such as net present value (NPV) and payback period in addition to the Modified Internal Rate of Return (MIRR). While the MIRR provides valuable insights into the potential profitability of an investment, incorporating these additional metrics can provide a more comprehensive analysis.
### Net Present Value (NPV)
The Net Present Value (NPV) is a financial metric that measures the present value of all cash flows related to an investment. It helps determine the value of an investment by discounting future cash flows to their present value. A positive NPV indicates that the investment is expected to generate more cash inflows than the initial investment cost.
Calculating NPV involves selecting a discount rate, which represents the minimum acceptable rate of return required by the investor. The cash flows, both positive and negative, are then discounted to their present value using this rate. The NPV is calculated by summing all the present values of the cash flows. A positive NPV suggests that the investment is expected to create value and is considered favorable.
### Payback Period
The payback period is a straightforward financial metric that measures the time taken for an investment to recoup its initial cost. It determines how quickly an investor can recover their investment. The payback period is especially useful for investors who prioritize short-term liquidity and want to assess the risk associated with recouping their initial investment in a timely manner.
The payback period is calculated by dividing the initial investment by the annual cash inflows. This provides the number of years required to recover the initial investment. A shorter payback period is usually preferred, as it indicates a quicker return on investment. However, it’s essential to consider other factors alongside the payback period, such as the profitability and sustainability of the investment in the long run.
Financial Metric Description
Modified Internal Rate of Return (MIRR) A financial metric that considers the reinvestment of cash flows and provides a more accurate assessment of an investment’s value and potential return.
Net Present Value (NPV) The present value of all cash flows related to an investment, helping determine its value and profitability.
Payback Period The time taken for an investment to recoup its initial cost, providing insights into the speed of return on investment.
## The Value of the MIRR Calculator
The MIRR calculator is a valuable financial analysis tool that simplifies complex calculations and helps investors make more informed decisions. By incorporating the reinvestment of cash flows at a separate rate of return, the MIRR provides a more accurate assessment of investment profitability. This powerful metric is especially useful for evaluating projects with intricate cash flow patterns and those that require significant financing.
With the MIRR calculator, users can input the initial investment, cash inflows, and cash outflows for each period, as well as the discount rate for reinvestment and financing. The calculator then calculates the MIRR, allowing investors to determine whether an investment is expected to generate a profit. If the calculated MIRR is higher than the required rate of return, it indicates a profitable opportunity.
In addition to its ability to evaluate investment profitability, the MIRR calculator offers other advantages in financial analysis. It simplifies complex calculations, saving time and effort for investors. Moreover, by considering the reinvestment of cash flows, the MIRR provides a more realistic assessment of investment value compared to traditional metrics like the Internal Rate of Return (IRR).
When assessing potential investment opportunities, it is essential to consider multiple financial metrics alongside the MIRR. Metrics such as net present value (NPV) and payback period provide a comprehensive evaluation of an investment’s viability. By combining these metrics, investors can gain a more comprehensive understanding of the risks and returns associated with an investment.
### Table: Comparison of Financial Metrics
MIRR Modified Internal Rate of Return – Considers reinvestment of cash flows
IRR Internal Rate of Return – Does not consider reinvestment of cash flows
NPV Net Present Value – Measures investment’s value in today’s dollars
Payback Period Time required to recoup initial investment – Provides quicker assessment of investment recovery
Overall, the MIRR calculator is an indispensable tool for investors seeking to make smart investment decisions. By providing a more realistic assessment of investment profitability and streamlining complex calculations, it empowers users to evaluate opportunities with greater accuracy and confidence. To maximize investment returns and mitigate risks, leverage the power of the MIRR calculator for comprehensive financial analysis.
## Applying the MIRR Calculator to Shrewd Investments
By applying the MIRR calculator to investment opportunities, individuals can make shrewd investment decisions and maximize their returns. The calculator allows users to assess the profitability of potential investments by considering the reinvestment of cash flows. This is particularly valuable for projects with complex cash flow patterns or those that involve substantial financing.
One of the key advantages of using the MIRR calculator is its ability to provide a more accurate assessment of investment value. Unlike the traditional IRR, which assumes that cash flows are reinvested at the same rate of return, the MIRR takes into account the actual rate at which cash inflows are reinvested. This consideration significantly impacts the evaluation of investment opportunities, as it reflects the real-world scenario of reinvesting cash flows at a separate rate.
Moreover, the MIRR calculator simplifies the process of evaluating investment profitability. By inputting the initial investment, cash inflows, cash outflows, and discount rate for reinvestment and financing, users can quickly calculate the MIRR. If the calculated MIRR is higher than the required rate of return, it indicates that the investment is expected to generate a profit. This information empowers individuals to make informed decisions and prioritize investments with the potential for higher returns.
Benefits of Using the MIRR Calculator
Accurate assessment of investment value
Consideration of cash flow reinvestment
Efficient evaluation of investment profitability
When considering investment opportunities, it is important to remember that the MIRR is just one metric to be taken into account. Alongside the MIRR, other financial metrics like net present value (NPV) and payback period should also be considered. Utilizing these metrics in combination provides a comprehensive understanding of investment potential.
In conclusion, the MIRR calculator is a powerful tool for evaluating investment opportunities and making shrewd investment decisions. By considering the reinvestment of cash flows, the MIRR provides a more realistic assessment of investment value and potential profitability. Take advantage of this specialized financial analysis tool to maximize your returns and achieve your investment goals.
## Conclusion
The MIRR calculator is a powerful financial analysis tool that enables individuals to make informed investment decisions and simplify their financial planning. By calculating the Modified Internal Rate of Return, the MIRR takes into account the reinvestment of cash flows at a separate rate of return, providing a more realistic assessment of investment opportunities. With the ability to input the initial investment, cash inflows, and cash outflows for each period, as well as the discount rate for reinvestment and financing, the MIRR calculator streamlines the evaluation process.
When the calculated MIRR is higher than the required rate of return, it suggests that the investment is expected to generate a profit. This metric is particularly valuable for assessing investments with complex cash flows or those that require significant financing. By providing a more accurate assessment of investment value and potential return, the MIRR calculator equips investors with the tools they need to make informed decisions.
While the MIRR is a powerful metric, it is essential to consider other financial evaluations such as net present value (NPV) and payback period alongside it. These additional metrics provide a comprehensive understanding of investment opportunities and further enhance decision-making. By utilizing the MIRR calculator alongside other financial tools, individuals can gain a holistic perspective on the profitability and viability of potential investments.
In conclusion, the MIRR calculator is an invaluable resource for individuals seeking to make informed financial decisions and evaluate investment opportunities. With its ability to consider the reinvestment of cash flows, this financial analysis tool provides a more accurate assessment of investment value. By leveraging the MIRR calculator, individuals can simplify their financial planning and maximize their investment returns.
## Get Started with the MIRR Calculator Today
Don’t miss out on the benefits of the MIRR calculator – start using this valuable financial analysis tool today for evaluating investments and making informed financial decisions. The MIRR (Modified Internal Rate of Return) calculator is a powerful metric that takes into account the reinvestment of cash flows at a separate rate of return, providing a more realistic assessment of investment opportunities.
With the MIRR calculator, you can input the initial investment, cash inflows, and cash outflows for each period, along with the discount rate for reinvestment and financing. The calculator then calculates the MIRR, allowing you to determine if the investment is expected to generate a profit.
The MIRR calculation is particularly useful for evaluating investments with complex cash flows or projects that require significant financing. By considering the reinvestment of cash flows, the MIRR provides a more accurate assessment of the investment’s value and potential return. When evaluating investment opportunities, it is also important to consider other metrics such as net present value (NPV) and payback period.
Start using the MIRR calculator today and make more informed financial decisions. Whether you’re a seasoned investor or just starting out, this valuable tool will help you evaluate investments accurately and maximize your returns. Don’t wait – take advantage of the MIRR calculator and unlock the power of financial analysis for investment evaluation.
## FAQ
### What is the MIRR calculator?
The MIRR calculator is a powerful financial tool that helps evaluate investment opportunities by calculating the rate at which cash inflows are reinvested at a separate rate of return.
### How is MIRR different from IRR?
Unlike the traditional Internal Rate of Return (IRR), which does not consider reinvestment of cash flows, the MIRR takes into account the reinvestment of cash flows at a separate rate of return, providing a more realistic assessment of investment opportunities.
### What inputs does the MIRR calculator require?
The MIRR calculator requires inputs such as the initial investment, cash inflows, cash outflows for each period, and the discount rate for reinvestment and financing.
### How does the MIRR calculator work?
The MIRR calculator calculates the MIRR based on the inputs provided. If the calculated MIRR is higher than the required rate of return, the investment is expected to generate a profit.
### What are the advantages of using the MIRR calculation?
The MIRR calculation provides a more accurate assessment of investment value by considering the reinvestment of cash flows. It helps investors make more informed decisions and evaluate the profitability of investments.
### When is the MIRR calculator particularly useful?
The MIRR calculator is particularly useful for evaluating investments with complex cash flows or projects that require significant financing. It aids in assessing the value and potential return of such investments.
### Should I consider other financial metrics alongside the MIRR?
Yes, it is important to consider other financial metrics such as net present value (NPV) and payback period alongside the MIRR to gain a comprehensive understanding of investment opportunities.
### How can I apply the MIRR calculator to make shrewd investments?
The MIRR calculator enables you to evaluate potential investments accurately, considering the reinvestment of cash flows. By utilizing the calculator, you can maximize your investment returns and make shrewd investment decisions.
### How can I get started with the MIRR calculator?
You can get started with the MIRR calculator today by inputting the necessary data such as initial investment, cash inflows, cash outflows, and the discount rate for reinvestment and financing. Use the calculator to evaluate investment opportunities and make smarter financial decisions.
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https://www.physicsforums.com/threads/find-dy-dx-as-a-function-of-t-for-the-parametric-equations.708788/
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# Find dy/dx as a function of t for the parametric equations
## Homework Statement
find dy/dx as a function of t for the parametric equations
x=cos^7(t)
y=6sin^2(t)
## The Attempt at a Solution
well i'm looking for dy/dx.. so first i found dy
dy = 12sin(t)cos(t)
and dx
dx = -7cos^6(t)sin(t)
dy/dx = 12sin(t)cos(t) / -7cos^6(t)sin(t)
(-12/7cos^5(t))
And then it asks for if the curve is concave up or concave down (besides where it is undefined). When i graph it on wolfram alpha it gives both up and down concaves so it leads me to believe I did this wrong. Help
CompuChip
Homework Helper
dy/dx is the derivative of y with respect to x, it is not technically a fraction. In many cases you can treat it as such, but this is one of those cases where it doesn't really work.
Does the chain rule ring a bell when written as
$$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt}$$
or, more conveniently for you,
$$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \text{ ?}$$
eumyang
Homework Helper
To find the concavity you need the second derivative. I only see the first derivative in your post. Also, when you say that you found "dy" and "dx", you should say instead "dy/dt" and "dx/dt".
EDIT: Regarding my second point, CompuChip beat me to it. :tongue:
dy/dx is the derivative of y with respect to x, it is not technically a fraction. In many cases you can treat it as such, but this is one of those cases where it doesn't really work.
Does the chain rule ring a bell when written as
$$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt}$$
or, more conveniently for you,
$$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \text{ ?}$$
It does not. I just started calculas 3 and we are going over alot of calculus one stuff which i got a D in over 5 years ago. lol ;-(. Calc 2 i stepped up my game and got an A but alot of this stuff is confusing me. I'll go relearn the chain rule. That dt is some kind of new variable that comes out of using the chain rule?
dy/dx = (dy/dt) / (dx/dt). I don't understand how this will get me a result different from what I did the first time. Isn't dy/dt 12sin(t)cos(t)? and dx/dt is 7cos^6(t)sin(t)?
eumyang
Homework Helper
dy/dx = (dy/dt) / (dx/dt). I don't understand how this will get me a result different from what I did the first time. Isn't dy/dt 12sin(t)cos(t)? and dx/dt is 7cos^6(t)sin(t)?
I didn't say that your first derivative was wrong (although you forgot a negative in dx/dt in your more recent post). I'm saying that to test for concavity you need to find the second derivative.
1 person
pbuk
Gold Member
dy/dx = (dy/dt) / (dx/dt). I don't understand how this will get me a result different from what I did the first time.
It won't - in this problem. But if you start writing things like dy = 12sin(t)cos(t)
1. it looks like you don't know what you are doing
2. when problems get harder (like in the second half of this question) you will get caught out
As already stated you need to look at ## \frac{d^2y}{dx^2} ## to get the curve's concavity.
okay.
so dy/dx = -12/(7cos^5(t))
I need to take the derivative again to find the concavity
(d/dt)(dy/dx) = (d/dt)(-12/(7cos^5(t)) = (-60/7)tan(t)sec^5(t). And yet when i graph this curve, it looks like it has sections where it concaves up and some where in concaves down.
alright so i see my above post is wrong.. Now i have
(d/dt)(dy/dx) / (dx/dt) = the second derivative i'm looking for.. after doing all the math i got (-60/49)sec^12(t).. anyone have advice?
I GOT IT thanks everyone :D
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https://www.techrepublic.com/article/how-to-change-the-axis-of-rotation-sort-of-for-a-shape-in-powerpoint/
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It’s easy to make shapes spin in Microsoft PowerPoint; simply apply the Spin animation. However, achieving the right spin within the right area isn’t always that simple. Sometimes you need to force the axis of rotation to recognize a different area—you can’t really do that, but there is a workaround that does this trick. In this article, I’ll show you how to do so by forcing a triangle to span the entire circumference of a circle, similar to the hands of a clock, instead of spinning in only one half of the circle.
SEE: Office 365: A guide for tech and business leaders (free PDF) (TechRepublic)
This article assumes you have basic PowerPoint skills, such as inserting shapes and applying properties. For your convenience, you can download the demonstration .pptx file. You can apply this workaround in the browser edition.
## What is axis of rotation?
Merriam Webster defines axis of rotation as follows: The straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles.
A great example of this concept is a compass dial. Its axis of rotation is the center of the dial. The dial circles around the circumference as expected because the axis of rotation is the center of the dial, which is also the center of the circle. This concept might be better explained with a visual.
## How to create a spinning arrow
A center axis of rotation is inherent in some shapes, such as lines. When you can use one of these shapes, do so, because what you need is already there. For example, let’s add a double-arrow shape and apply a spin animation to create a simple spinning arrow:
1. Click the Insert tab, click Shapes in the Illustrations group, and choose the left to right double-arrow shape in the Block Arrows section. Position and size the inserted arrow (Figure A).
2. Click the Animations tab.
3. From the Emphasis section, choose Spin. PowerPoint will give you a quick preview.
Figure A
A figure can’t show you the full spin, but the arrow spins from the center—its axis of rotation is in the center of the arrow. (Figure B).
Figure B
Play around with a few different shapes and you’ll see that many shapes spin as expected, whereas others don’t. For instance, try a square and a triangle, and you’ll find that the triangle spins, but the axis of rotation might not be where you expect!
## Creating a false axis of rotation
It would be great if every shape spun as expected, but because of the axis of rotation, some will surprise you. We can illustrate this by trying to create a clock dial. The double-arrow will spin correctly, but it’s a double-arrow, not a traditional one-arrow dial.
To get the results you want, you might use a long, narrow triangle, as shown in Figure C. But the lone triangle won’t spin as expected. Instead of spinning around the circle, which is what you want, it spins in the top half only (see Figure D). That’s because it exists only in the top half. It’s spinning correctly, but it’s not what you want.
Figure C
Figure D
The obvious solution is to change the axis of rotation to the bottom of the triangle. There’s no way to do that, but there is a workaround. In a nutshell, copy the triangle and rotate it, lining it up with the bottom to create a double-arrow dial, then set the new triangle’s fill property to nothing. Finally, you group the two shapes and in doing so, you change the axis of rotation from the center of the lone triangle to the point where the two arrows meet.
Let’s work through an example using the triangle in Figure C (in the downloadable demonstration file):
1. After positioning and sizing the narrow triangle, select it, press Ctrl+B and drag a copy to the side.
2. With the copy still selected, choose Rotate from the Arrange dropdown (on the Home tab).
3. From the submenu, choose Flip Vertical (Figure E).
4. Drag the copy to the lower half and line it up with the arrow in the top half (Figure E).
5. With the lower triangle still selected, choose No Fill from the Shape Fill dropdown (Shape Styles group on the contextual Shape Format tab).
6. Choose No Outline from the Shape Outline dropdown.
7. Hold down the shift key and select both triangles and then press Ctrl+G to group the two. (If you’re working in the browser editing, hold down the Shift key.) The two selection boxes will turn into one. If this doesn’t happen, try again.
8. With the grouped triangles selected, apply the Spin animation as you did before.
Figure E
By changing the axis of rotation from the center of the upper triangle to the center of the fake dial, you force the spin animation to work around the clock! As you can see in Figure F, the upper triangle now spins through both the upper and lower halves of the circle. You can use this workaround any time you need to adjust the axis of rotation for a shape.
Figure F
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# Why data analysts should exercise caution when taking averages (Part 2)
This is Part 1 in a 2-part series on miscalculating averages and its implications.
In Part 1 of this series, I explained the fundamental problem with calculating multidimensional averages: you can never be sure what a single observation represents in the database (given by the primary key), and as a result, you need to "guarantee" what an observation is by specifying a collapsing key. While Part 1 emphasized how to structurally think about the problem of multidimensional averages, this post covers some of its implications.
## 1. When it comes to averages, what makes a good question?
While every analysis begins with a good question, parsing that question linguistically can sometimes be challenging.
As a recap of my first post, every question with an average should contain at least two elements and sometimes three:
1. Measure
2. Averaging key (specifies the denominator)
3. Grouping key (optional)
For example, "what is the average daily quantity of products sold?" has a measure of "quantity" and an averaging key of "day". Similarly, "How many texts does the average person send per day?" has a measure of "number of texts", a grouping key of "day" and an averaging key of "person". I think it often helps to rephrase the question like this: "the average [measure] across [averaging key] by [grouping key]."
In contrast, "what is the daily average?" is an incomplete question because there is no measure. Similarly, "what is the average quantity purchased" is incomplete because there is no averaging key (that is, average quantity purchased across people, across cohorts, over time, etc.?).
Thinking in terms of grouping and averaging keys can help parse linguistically similar questions:
• "What is the average daily share price over the last 20 days?"
• Output: a single value, the average, calculated by summing each day's closing share price over the last 20 days divided by 20.
• "What is the average share price by day for the last 20 days?
• Output: a list of 20 days (ie. "for each day"), each with its average share price over the day. For example, if your data comes at the minute-level (ie. averaging key is "across minutes"), each average would be sum of all prices that trading day divided by the count of minutes.
• "What is the average daily share price by security in a portfolio over the last 20 days?
• Output: a list of securities, (ie. "for each security"), each with a daily average over the last 20 days.
• "What is the average share price by security by day?"
• Output: [ambiguous - grouping key and averaging key are indiscernible]
To reiterate, the key here is knowing that the averaging key defines how many records will be in the denominator. If you are unsure what your denominator is, or conversely, how to interpret your averaging key, you will have trouble answering your initial question.
## 2. What aggregation functions besides the mean are susceptible to the pitfall?
Unfortunately, while the difficulty of working with multidimensional data may be most salient with the arithmetic mean, it is not confined to it. In fact, many other aggregation functions, such as the standard deviation (also uses n as the denominator), median, mode, maximum, minimum, first, last, count and count may all be similarly affected.
We can see this with an example.
Example 1
date product state purchaser quantity transactions 2016-12-23 vacuum NY Brian King 1 2016-12-23 stapler NY Brian King 3 2016-12-23 printer ink NY Brian King 2 2016-12-23 vacuum MA Lauren Mills 1 2016-12-23 printer ink MA John Smith 5 2016-12-24 vacuum MA Lauren Mills 1 2016-12-24 keyboard NY Brian King 2 2016-12-24 stapler NY Trevor Campbell 1 2016-12-25 keyboard MA Tom Lewis 4
"For each day, what is the max quantity purchased by any individual?"
Since the primary key is more granular than the key on which we are analyzing - that is, (purchaser) - we again will have to collapse the data. In SQL:
```SELECT date, MAX(quantity) FROM (
SELECT date, purchaser, SUM(quantity)
FROM transactions
GROUP BY date, purchaser // COLLAPSING KEY
) AS t
GROUP BY date // GROUPING KEY
// MAX KEY (IMPLIED): [date, purchaser] - [date] = [purchaser]```
From this example, it's clear that while there is no denominator to be affected, aggregating the actual values themselves changes what the answer will be. We can imagine how median, mode and the other calculations are all similarly affected when the underlying data is aggregated into new values.
## 3. Can you interpret an averaging key with more than one dimension?
Yes, the averaging key can be a composite key (ie. composed of one or more dimensions). However, the composite key has to make sense.
Example 2
first_name last_name num_apples apples Katie Smith 4 Alan Johnson 8 Alan Howard 10 Tess O'Neal 8 Katie Powers 5
For example, the composite key of (first_name, last_name) makes sense because it has a natural interpretation: full_name. Full name corresponds to a single individual, and it makes sense to take averages on the level of individuals - as a result, (first_name, last_name) would be a legitimate averaging key. Many times, however, the averaging key will not make sense with two or more dimensions - this is an indication that more aggregation (ie. the collapsing key) should have been performed.
## 4. How far off can the calculated average be with the wrong averaging key?
By way of example, we can evaluate how the calculated average directionally changes as we change the length of the averaging key.
Looking at Example 2 above, what is the average when the averaging key is (first_name, last_name)? It is SUM(4 + 8 + 10 + 8 + 5) / 5 = 35. What is the average when the averaging key is (first_name)? It is SUM(4 + 8 + 10 + 8 + 5) / 3 = 11.66.
As you can tell, the more granular the the averaging key, the more records will be in the denominator (ie. higher) and the smaller the average will be. Conversely, the more collapsed the averaging key, the fewer records will be in the denominator (ie. lower) and the larger the average will be. For highly dimensional datasets (more granularity), this means averages can be significantly lower than what makes sense due to inadequate aggregation.
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# Magnetic Flux Density B
Magnets exert attractive or repulsive force. This is called magnetism. The magnetic effect is higher in regions closer to the magnetic poles and reduces as one move farther away from the magnetic. The region in which magnetic force is exerted is called a magnetic field.
Magnetic fields can be represented using vectors. This is because these fields have a magnitude and direction.Magnetic Field 1Magnetic Field 1Types of Magnetic flux density BThere are two types of vectors that represent magnetic fields,One is magnetic induction also called as magnetic flux density represented by B and the other is magnetic field strength represented by H.in vacuum, B and H cannot be distinguished.
The magnetic flux density in the B field depends on macroscopic and microscopic current based on the electron movement in the nuclei. Therefore magnetic flux density can be defined in several ways based on the effect of flux density on the environment.Consider a charged particle p moving in a magnetic field with a velocity equal to v. this charge experience a force in the magnetic field. Thisforce can be represented asF = pv *BHere p is the charge of the particlev is the velocity and B is the magnetic field.magnetic flux density can also be defined as equivalent to the torque on a magnetic dipole placed in a magnetic field.
The magnetic flux density is representedusing teslas in SI units andUsing gauss in cgs units. (1 tesla = 10,000 gauss).The SI unit of tesla = (newton × second)/ (coulomb × meter)Magnetic flux density can be measured using magnetometers.The other names for magnetic flux density as used in different industries ismagnetic flux density used by electrical engineers, magnetic induction used by applied mathematicians and electronics engineers and magnetic field used by physicists.Conclusion for
Between, if you have problem on these topics exponential decay equation, please browse expert math related websites for more help on Escape Velocity Formula.
Magnetic flux density BMagnetic flux density depends on macroscopic as well as microscopic current created by electron movement in the nucleus. Magnetic flux density is the term used for magnetic field in B field and opposed to magnetic field in H field that is called magnetic field strength.
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https://www.teacherspayteachers.com/Product/Addition-within-1000-and-Subtraction-within-1000-Math-Activity-3514863
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# Addition within 1000 and Subtraction within 1000 Math Activity
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10 MB|30 pages
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1. Put some fun in your classroom with these no prep activities that will allow your students to work on their fraction skills. This bundle includes five of my fraction review games.→What skills does this mystery focus on?These products cover a wide variety of 4th grade skills, including fractions, add
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Product Description
Put some fun in your classroom with this no prep math activity that will allow your students to work on their addition and subtraction within 1000 skills. This math game is a perfect center activity that will give your students practice in solving word problems and standard equations.
→What skills does this mystery focus on?
This product focuses on addition and subtraction within 1000. This targets both word problems and standard equations. Each problem is presented with 3 options for students to choose from as their correct answer.
→What mystery are students working to solve?
This is a birthday party math mystery. In this product, students are given the names of 5 children. Each child has brought a gift to a birthday party; but no one can remember who brought which gift and in what order they were opened. Help the friends determine which child (Marcia, Dwayne, Annette, Jack, Sam) brought which gift (fishing pole, books, art kit, science kit, basketball) and what order it was opened (first, second, third, fourth, fifth).
→What is a Mystery Product?
Mystery products are part of my newest product line. Mysteries are a great way for students to practice a skill with a fun twist. Students will complete 20 multiple choice questions. After each question, they will receive a clue based on their answer. After they have correctly answered each of the questions, students will have solved the mystery.
→What makes this product so great?
These are no prep! (Unless you want to cut the 10 pages of questions in half.)
Students can look at the questions in any order. They can rotate through as quickly (or as slowly) as they need to. This makes management a breeze!
Tracking on the bottom allows students to color in a square as they solve, meaning they won't be confused on which problems they haven't figured out yet.
A full answer key, means you will be able to quickly determine if/where an error was made!
Sometimes students make a mistake, and that's ok too. These are also self checking. Students will receive clues immediately; so if they receive a clue that contradicts a previous clue the student will know know he has made a mistake.
→Do students have to work with a partner?
Nope! That's up to you! These can be used for independent practice, partner work, or group work. Perhaps you'll give a couple of clues a day which will spread the activity out for the entire week! Use this as a morning work activity, or a fast finisher choice!
→What's included?
Setup Directions for Teachers
An introduction story to begin the mystery
10 Pages of Problems (20 problems in all)
2 Recording sheet for students
A full answer key that mimics the 10 question pages and recording sheets
→I love this product! Do you have any more?
Yes! This is my newest product line, and I'm continuing to add more topics. If you have an idea, feel free to send me an email at giraffic.jam.tpt@gmail.com or by using the "Ask a Feature" section on TPT. To see my other products like this one, CLICK HERE
Get this product as part of my Mystery Math Bundle here.
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# The Azimuth Project Blog - network theory (part 14)
This page is a blog article in progress, written by John Baez. To see discussions of it as it was being written, go to the Azimuth Forum. For the final polished version, visit the Azimuth Blog.
We’ve been doing a lot of hard work lately. Let’s take a break and think about a fun example from chemistry!
#### The ethyl cation
Suppose you start with a molecule of ethane, which has 2 carbons and 6 hydrogens arranged like this:
Then suppose you remove one hydrogen. The result is a positively charged ion, or ‘cation’. This particular one is called an ‘ethyl cation’. People used to think it looked like this:
They also thought a hydrogen could hop from the carbon with 3 hydrogens attached to it to the carbon with 2. So, they drew a graph with a vertex for each way the hydrogens could be arranged, and an edge for each hop. It looks really cool:
The red vertices come from arrangements where the first carbon has 2 hydrogens attached to it, and the blue vertices come from those where the second carbon has 2 hydrogens attached to it. So, each edge goes between a red vertex and a blue vertex.
This graph has 20 vertices, which are arrangements or ‘states’ of the ethyl cation. It has 30 edges, which are hops or ‘transitions’. Let’s see why those numbers are right.
First I need to explain the rules of the game. The rules say that the 2 carbon atoms are distinguishable: there’s a ‘first’ one and a ‘second’ one. The 5 hydrogen atoms are also distinguishable. But, all we care about is which carbon atom each hydrogen is bonded to: we don’t care about further details of its location. And we require that 2 of the hydrogens are bonded to one carbon, and 3 to the other.
If you’re a physicist, you may wonder why the rules work this way: after all, at a fundamental level, identical particles aren’t really distinguishable. I’m afraid I can’t give a fully convincing explanation right now: I’m just reporting the rules as they were told to me!
Given these rules, there are 2 choices of which carbon has two hydrogens attached to it. Then there are $\binom{5}{2} = 10$ choices of which two hydrogens are attached to it. This gives a total of $2 \times 10 = 20$ states. These are the vertices of our graph: 10 red and 10 blue.
The edges of the graph are transitions between states. The idea is that any hydrogen in the group of 3 can hop over to the group of 2. There are 3 choices for which hydrogen atom makes the jump. So, starting from any vertex in the graph there are 3 edges. This means there are $3 \times 20 / 2 = 30$ edges. Why divide by 2? Because there are 3 edges touching each of 20 vertices, but each edge actually touches 2 vertices, so we must divide by 2 to avoid double-counting them.
#### The Desargues graph
The idea of using this graph in chemistry goes back to this paper:
• A. T. Balaban, D. Fǎrcaşiu and R. Bǎnicǎ, Graphs of multiple 1,2-shifts in carbonium ions and related systems, Rev. Roum. Chim. 11 (1966), 1205.
This paper is famous because it was the first to use graphs in chemistry to describe molecular transitions, as opposed to using them as pictures of molecules!
But this particular graph was already famous for other reasons. It’s called the Desargues-Levi graph, or Desargues graph for short:
Desargues graph, Wikipedia.
Later I’ll say why it’s called this.
There are lots of nice ways to draw the Desargues graph. For example:
The reason why we can draw such pretty pictures is that the Desargues graph is very symmetrical. Clearly any permutation of the 5 hydrogens acts as a symmetry of the graph, and so does any permutation of the 2 carbons. This gives a symmetry group $S_5 \times S_2$, which has $5! \times 2! = 240$ elements. And in fact this turns out to be the full symmetry group of the Desargues graph.
The Desargues graph, its symmetry group, and its applications to chemistry are discussed here:
• Milan Randić, Symmetry properties of graphs of interest in chemistry: II: Desargues-Levi graph, Int. Jour. Quantum Chem. 15 (1997), 663-682.
#### The ethyl cation, revisited
We can try to describe the ethyl cation using probability theory. If at any moment its state corresponds to some vertex of the Desargues graph, and it hops randomly along edges as time goes by, it will trace out a random walk on the Desargues graph. This is a nice example of a Markov process!
We could also try to describe the ethyl cation using quantum mechanics. Then, instead of having a probability of hopping along an edge, it has an amplitude of doing so. But as we’ve seen, a lot of similar math will still apply.
It should be fun to compare the two approaches. But I bet you’re wondering which approach is correct. This is a somewhat tricky question, at least for me. The answer would seem to depend on how much the ethyl cation is interacting with its environment—for example, bouncing off other molecules. When a system is interacting a lot with its environment, a probabilistic approach seems to be more appropriate. The relevant buzzword is ‘environmentally induced decoherence’.
However, there’s something much more basic I have tell you about.
After the paper by Balaban, Fǎrcaşiu and Bǎnicǎ came out, people gradually realized that the ethyl cation doesn’t really look like the drawing I showed you! In fact it’s what chemists call ‘nonclassical’ ion. What they mean is this: its actual structure is not what you get by taking the traditional ball-and-stick model of an ethane molecule and ripping off a hydrogen. The ethyl cation really looks like this:
For more details, and pictures that you can actually rotate, see:
• Stephen Bacharach, Ethyl cation, Computational Organic Chemistry.
So, if we stubbornly insist on applying the Desargues graph to realistic chemistry, we need to find some other molecule to apply it to.
#### Trigonal bipyramidal molecules
Luckily, there are lots of options! They’re called trigonal bipyramidal molecules. They look like this:
The 5 balls on the outside are called ‘ligands’: they could be atoms or bunches of atoms. In chemistry, ‘ligand’ just means something that’s stuck onto a central thing. For example, in phosphorus pentachloride the ligands are chlorine atoms, all attached to a central phosphorus atom:
It’s a colorless solid, but as you might expect, it’s pretty nasty stuff: it’s not flammable, but it reacts with water or heat to produce toxic chemicals like hydrogen chloride.
Another example is iron pentacarbonyl, where 5 carbon-oxygen ligands are attached to a central iron atom:
You can make this stuff by letting powdered iron react with carbon monoxide. It’s a straw-colored liquid with a pungent smell!
Whenever you’ve got a molecule of this shape, the ligands come in two kinds. There are the 2 ‘axial’ ones, and the 3 ‘equatorial’ ones:
And the molecule has $20$ states… but only if count the states a certain way. We have to treat all 5 ligands as distinguishable, but think of two arrangements of them as the same if we can rotate one to get the other. The trigonal bipyramid has a rotational symmetry group with 6 elements. So, there are $5! / 6 = 20$ states.
The transitions between states are devilishly tricky. They’re called pseudorotations, and they look like this:
If you look very carefully, you’ll see what’s going on. First the 2 axial ligands move towards each other to become equatorial.
Now the equatorial ones are no longer in the horizontal plane: they’re in the plane facing us! Then 2 of the 3 equatorial ones swing out to become axial. This fancy dance is called the Berry pseudorotation mechanism.
To get from one state to another this way, we have to pick 2 of the 3 equatorial ligands to swing out and become axial. There are 3 choices here. So, if we draw a graph with states as vertices and transitions as edges, it will have 20 vertices and $20 \times 3 / 2 = 30$ edges. That sounds suspiciously like the Desargues graph!
Puzzle 1. Show that the graph with states of a trigonal bipyramidal molecule as vertices and pseudorotations as edges is indeed the Desargues graph.
I think this fact was first noticed here:
• Paul C. Lauterbur and Fausto Ramirez, Pseudorotation in trigonal-bipyramidal molecules, J. Am. Chem. Soc. 90 (1968), 6722–6726.
Okay, enough for now! Next time I’ll say more about the Markov process or quantum process corresponding to a random walk on the Desargues graph. But since the Berry pseudorotation mechanism is so hard to visualize, I’ll pretend that the ethyl cation looks like this:
and I’ll use this picture to help us think about the Desargues graph. That’s okay: everything we’ll figure out can easily be translated to apply to the real-world situation of a trigonal bipyramidal molecule. The virtue of math is that when two situations are ‘mathematically the same’, or ‘isomorphic’, we can talk about either one, and the results automatically apply to the other. This is true even if the one we choose to talk about doesn’t actually exist in the real world!
category: blog
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Physics Assignment
I. Problems with Electrical Current
1. If there is a current of 10 amperes in a circuit for 10 minutes, what quantity of electric charge flows in through the circuit?
2. How much current must there be in a circuit if 100 coulombs flow past a point in the circuit in 4 seconds?
3. How much time is required for 10 coulombs of charge to flow past a point if the rate of flow (current) is 2 amperes?
II. Problems with Electrical Potential
4. The potential difference between two point A and B in an electric field is 25 volts. How much work is required to transfer 10 coulombs of charge from A to B?
5. Ten joules of work are required to transfer 2 coulombs of charge from X to Y. What is the difference in potential between these two points?
6. It requires 600 joules of energy to transfer a quantity of charge between points C and D at a potential difference of 30 volts. How much charge is transferred?
III. Problems with Current, Potential, and Power
7. If 100 volts causes a current of 5 amperes in a lamp for 20 seconds,
(A) What quantity of charge flows through the lamp?
(B) How much energy is used?
(C) What power is developed?
8. The potential difference between two points is 100 volts.
(A) What quantity of charge would be transferred by the use of 400 joules in 2 seconds?
(B) What current would exist in the circuit?
(C) What power would be developed in the circuit?
9. The potential difference of a battery is 25 volts and 10 amperes flow in the circuit.
(A) What quantity of charge flows in the circuit in 20 seconds?
(B) How much energy is delivered?
(C) What power is developed?
10. One hundred joules is used to transfer 200 coulombs from points A to B in 5 seconds.
(A) How much current exists in the circuit?
(B) What is the potential difference between the two points?
(C) What power is needed for this transfer?
IV. Problems in Conversions
11. How many kilowatts is 250 watts?
12. How many kilowatts is 1.0 x 105 watts?
13. How many joules are in 746 watt-seconds?
14. How many joules are in one kilowatt-hour?
Back to the Brockport High School Science Department
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# 3/15/11
Written by G. T. Hushion. Posted in Articles
Understanding “action at a distance” is helped by looking closely at the corpse of the old crow, “the wheel has no memory.” A quick read of the page is linked in the “forum” section of this site.
Let a visual grasp of a yard stick with three poles replace the wheel of four poles….
…And understand that, relative to the serial random measurement of gravity …the 3 pole straight line of geometric probability is all that is that is rotating.
Let a random ball land in South. Let the second ball land anywhere. Predict the third ball to land in what we perceive as the Cardinal pole “North.” In geometric fact, at the third trial, the three part structure of the prediction of “bet” matches the 3 pole structure of the pi-angle …and at the third trial, the third pole naturally appears with a .33333 geometric probability, factored by two directions.
Since the “wheel” or “game” is only a mathematical perception relative to gravity, so too, “relative North” as a Cardinal pole is also only a perception.
With “action at a distance” naturally delivers a relative pi-angle pole at the third trial, it appears to contradict traditional random theory. Using Monte Carlo methodology, the “distance” between one random measurement and the next is 1/4 of the wheel between. It is the distance between each Cardinal pole. This allows North to only be predicted, relative to South, as a .25 algebraic possibility. That is: North is 1 of 4 possibilities. That is: under traditional random theory, North will be mathematically found at the third random trial only by also predicting or “betting” East or West.
When the finesse methodology of “action at a distance” is used on the wheel, East or West are allowed to algebraically occur, but are not predicted or bet. Relative to the random event in South, East and West are the “game” …and North is the far side of gravity.
With “action at a distance,” East and West disappear into cloud of algebra and North appears in the near distance as the geometric probability of the third pole on a pi-angle (or “diameter”) of three poles.
In short, the mechanics of “action at a distance” appear to mathematically shorten the distance, or number of occurrences, between opposing poles.
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https://www.physicsforums.com/threads/psi-of-water-in-a-cup.158054/
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# PSI of water in a cup
• Idea04
the bottom of the glass has atmospheric pressure because that is what is causing the pressure at that point (the weight of the air column above it)f
## Homework Statement
If you take a cup of water upside down on top of water in a bowl. The water doesn't flow out because of atmospheric force. So my question is, is the atmospheric pressure of 14.7psi transferred to the water in the cup. So is the water pressure inside the cup then 14.7 psi?
## The Attempt at a Solution
if the pressure inside the cup was not equal to the atmospheric pressure, what would happen?
So water open to the air will always have a pressure of 14.7psi. plus its own pressure.
So if the atmospheric pressure is transferred to the water, If you take a column of water 12 feet vertical with the top closed off and water entering at the bottom, with no air inside the column the water would rise to even pressure. So my question is would the atmospheric pressure be transferred throughout the water in the column. So at the top of the column would the pressure be 14.7 psi?
well, think about the units for stating a pressure as a height of a fluid
mmHg (milimeters of mercury)
760 mmHg = 14.7psi
that is, a column of mercury 760mm high will exert a pressure at the bottom, equal to atmospheric pressure
note that mercury is much denser than water and the water column that would provide a pressure of 14.7 psi would actually be 13.6 times taller
(because the density of water is 13.6 times that of mercury)
the pressure at the bottom of a column of liquid is due to its weight
it can be calculated from:
P = rho*g*h
rho being the density of the fluid, g being gravity, and h being the height of the column
So at the top of the column would the pressure be 14.7 psi?
err
at the top of the column...
no
the pressure that the column of water exerts is the weight of all the fluid
so at the top of the column it's just vapor pressure which is negligibly small
hmm...let me think about it a little more
i think it depends on the size of the glass
a small glass would not have a water column that would provide enough pressure to balance with the atmospheric pressure, so it seems that the pressure at the top would be the pressure needed to reach atmospheric after taking into account the pressure provided from the small column of water
alternatively...a glass that's 15 m tall, would be able to balance atmospheric pressure with the height of the water column it can provide, so the pressure at the top of the glass would basically be 0
Last edited:
okay, But having a column of water closed off at the top so no air can push downward is different then water sitting in a open top container.
The pressure of the atmosphere pushes the water up into the column. Keeping in mind that there is no air inside the column. So if the column was filled to the top it would exert around 5 psi downward pressure. That would be less than atmospheric pressure.
So my way of thinking, the pressure would be 14.7 psi in the column.
But would that be transferred to all parts of the column, or would the top of the column be 9.7psi and the bottom be 14.7 psi?
The pressure of the atmosphere pushes the water up into the column. Keeping in mind that there is no air inside the column. So if the column was filled to the top it would exert around 5 psi downward pressure. That would be less than atmospheric pressure.
So my way of thinking, the pressure would be 14.7 psi in the column.
But would that be transferred to all parts of the column, or would the top of the column be 9.7psi and the bottom be 14.7 psi?
if the glass was full, and the water in the bowl is trying to force its way in with atmospheric pressure
then the pressure at the top of the glass is the difference in pressure between atmospheric and the pressure provided by the water column inside the glass
I suppose you could consider the water column to provide negligible pressure and say that the pressure at the top of the glass is atmospheric pressure
it wouldn't be the exact answer, but a water column in a normal sized glass wouldn't provide very much pressure at all
the main thing, is that the pressure where the glass meets the free surface has to be atmospheric pressure
it'll be a constant at any point in that elevation
looking at the surface of the water in the bowl, it has atmospheric pressure at that elevation because that is what is causing the pressure at that point (the weight of the air column above it)
looking at the water at the bottom of the glass at the same elevation as the surface of the water in the bowl, it has atmospheric pressure at this elevation but it is not due to the column of air above it, it is due to the weight of the column of water above it and the pressure inside the glass
since no size of glass was given, it would not be wise to assume the pressure provided by the column of water is negligible
therefore,
the pressure at the top of the glass is the difference in pressure between atmospheric and the pressure provided by the column of water
you can apply bernoulli's equation to the top of the glass and the free surface in order to reach this conclusion as well
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https://m.lotterypost.com/thread/173988
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# What are the odds Pick 3 and 4 ? anybody - math wizards
Lee
This has probably already been discussed but what are the odds?
Pick 3 start 1000 to 1 straight with 1 draw - then with 4 predraws does this equal 5000 to 1 ?
Pick 4 start 10000 to 1 straight with 1 draw - then with 4 predraws does this equal 50,000 to 1 ?
Thanks
Lee
Lotterologist
The odds in the Pick 3 and Pick 4 never changes.
The odds in the Pick 4 are always 1 in 10,000 and the odds in the Pick 3 are always 1000 to 1.
Lee
Come on the odds increase when you add 4 predraws and more sets of balls and machines with rotation .
My picks hit in a predraw not in the 5th draw.
The odds are still 1000 to 1 and 10000 to 1 but they are manipulated with a purpose to make you lose not to see
if the machines are working properly. They increase the odds of prediction! I'm not just talking about the regular odds.
I know you don't have to play or lose money.
Thanks anyway
Lee
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https://www.bystudin.com/the-gardener-must-plant-6-trees-within-three-days-in-how-many-ways-can-he-distribute-the-work-over-the-days-if-he-plants-at-least-one-tree-a-day/
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# The gardener must plant 6 trees within three days. In how many ways can he distribute the work over the days if he plants at least one tree a day?
Suppose a gardener plants trees in a row and can make different decisions about which tree to stop after on the first day and after which on the second. Thus, one can imagine that the trees are separated by two partitions, each of which can stand in one of 5 places (between the trees). The partitions should be there one at a time, because otherwise not a single tree will be planted some day. Thus, you need to select 2 elements out of 5 (no repetitions). Therefore, the number of ways C2 5 = 10.
Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.
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tuts.r-nepal.org
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# Problem
We have data of medical insurance of patients. We will use the independent data to create a machine learning model which will estimate the Insurance charges. The medical charge is a numeric value so this problem is a regression problem.
Charge is dependent variable and these are independent variable:
• age: Integer indicating the age of the patients
• sex: patients gender ,either male or female
• bmi: Body Mass Index(BMI) ,BMI is equal to weight( in kilograms) divided by height(in meter) squared.It provides a sense of how over or under weight a person is relative to their height.
• children: An integer indicating the number of children/dependent covered by the insurance plan.
• smoker : Patients regularly smokes tobacco
• region : Patients place of residence in U.S
• charges: charge of the health insurance to patient yearly.
library(dplyr)
library(caret)
# Exploratory Data Analysis (EDA)
It is already done in data analysis section.
age sex bmi children smoker region charges
19 female 27.900 0 yes southwest 16884.924
39 male 33.770 1 no southeast 1725.552
28 male 33.000 3 no southeast 4449.462
33 male 22.705 0 no northwest 21984.471
32 male 28.880 0 no northwest 3866.855
31 female 25.740 0 no southeast 3756.622
str(insurance)
## 'data.frame': 1338 obs. of 7 variables:
## $age : int 19 39 28 33 32 31 46 37 37 60 ... ##$ sex : Factor w/ 2 levels "female","male": 1 2 2 2 2 1 1 1 2 1 ...
## $bmi : num 27.9 33.8 33 22.7 28.9 ... ##$ children: int 0 1 3 0 0 0 1 3 2 0 ...
## $smoker : Factor w/ 2 levels "no","yes": 2 1 1 1 1 1 1 1 1 1 ... ##$ region : Factor w/ 4 levels "northeast","northwest",..: 4 3 3 2 2 3 3 2 1 2 ...
## \$ charges : num 16885 1726 4449 21984 3867 ...
## Algorithm
As this problem is regression problem we will use Multiple Linear Regression Algorithm to make Medical insurance Predictive Model.
### Simple Linear Regression
simple linear regression is a simple method for predicting the quantitative value and study relationships between two continuous variables suppose X and Y. Mathematically, simple linear regression can be written as:
$Y=a+b∗X+e$
Where $$Y$$ is dependent variable, $$X$$ is independent variable, $$a$$ is the intercept , $$b$$ is the slope of $$X$$ and $$e$$ is the error term in equation.
Linear regression method’s main task is to find the best-fitting straight line through the Y and X points
### Multiple Linear Regression
Multiple linear regression attempts to model the relationship between two or more explanatory variables and a response variable by fitting a linear equation to observed data.
Multiple Linear regression uses multiple predictors. The equation for multiple linear regression looks like:
$Y = \beta0 + \beta1x1+ \beta2x2+ ...+e$
where:
$$Y$$ is Response or dependent variable $$\beta0$$ is intercept $$x1$$ and $$x2$$ are predictors or independent variable $$\beta1$$ and $$\beta2$$ are coefficeints for the $$x1$$ and $$x2$$ respectively and $$e$$ is error term in equation.
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https://infocenter-archive.sybase.com/help/topic/com.sybase.dc36271_36272_36273_36274_1250/html/refman/X48334.htm
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compute clause
Description
Generates summary values that appear as additional rows in the query results.
Syntax
```start_of_select_statement
compute row_aggregate (column_name)
[, row_aggregate(column_name)]...
[by column_name [, column_name]...]
```
Parameters
row_aggregate
is one of the following:
Function
Meaning
sum
Total of values in the (numeric) column
avg
Average of values in the (numeric) column
min
Lowest value in the column
max
Highest value in the column
count
Number of values in the column
column_name
is the name of a column. It must be enclosed in parentheses. Only numeric columns can be used with sum and avg.
One compute clause can apply several aggregate functions to the same set of grouping columns (see Examples 2 and 3). To create more than one group, use more than one compute clause (see Example 5).
by
calculates the row aggregate values for subgroups. Whenever the value of the by item changes, row aggregate values are generated. If you use by, you must use order by.
Listing more than one item after by breaks a group into subgroups and applies a function at each level of grouping.
Examples
Example 1
Calculates the sum of the prices of each type of cook book that costs more than \$12:
```select type, price
from titles
where price > \$12
and type like "%cook"
order by type, price
compute sum(price) by type
```
```type price
--------- ------------
mod_cook 19.99
sum
------------
19.99
type price
--------- ------------
sum
------------
35.94
(5 rows affected)
```
Example 2
Calculates the sum of the prices and advances for each type of cook book that costs more than \$12:
```select type, price, advance
from titles
where price > \$12
and type like "%cook"
order by type, price
```
```type price advance
--------- --------- ------------
mod_cook 19.99 0.00
sum sum
--------- ------------
19.99 0.00
```
```type price advance
--------- --------- ------------
sum sum
--------- ------------
35.94 15,000.00
(5 rows affected)
```
Example 3
Calculates the sum of the prices and maximum advances of each type of cook book that costs more than \$12:
```select type, price, advance
from titles
where price > \$12
and type like "%cook"
order by type, price
```
```type price advance
--------- --------- -------------
mod_cook 19.99 0.00
sum
---------
19.99
max
-------------
0.00
```
```type price advance
--------- --------- -------------
sum
---------
35.94
max
-------------
8,000.00
(5 rows affected)
```
Example 4
Breaks on type and pub_id and calculates the sum of the prices of psychology books by a combination of type and publisher ID:
```select type, pub_id, price
from titles
where price > \$10
and type = "psychology"
order by type, pub_id, price
compute sum(price) by type, pub_id
```
```type pub_id price
------------ --------- -----------
psychology 0736 10.95
psychology 0736 19.99
sum
---------
30.94
```
```type pub_id price
------------ --------- ---------
psychology 0877 21.59
sum
---------
21.59
(5 rows affected)
```
Example 5
Calculates the grand total of the prices of psychology books that cost more than \$10 in addition to calculating sums by type and pub_id:
```select type, pub_id, price
from titles
where price > \$10
and type = "psychology"
order by type, pub_id, price
compute sum(price) by type, pub_id
compute sum(price) by type
```
```type pub_id price
------------ --------- ---------
psychology 0736 10.95
psychology 0736 19.99
sum
---------
30.94
```
```type pub_id price
------------ --------- ---------
psychology 0877 21.59
sum
---------
21.59
sum
---------
52.53
(6 rows affected)
```
Example 6
Calculates the grand totals of the prices and advances of cook books that cost more than \$10:
```select type, price, advance
from titles
where price > \$10
and type like "%cook"
```
```type price advance
--------- ----------- --------------
mod_cook 19.99 0.00
sum sum
----------- --------------
67.88 19,000.00
(5 rows affected)
```
Example 7
Calculates the sum of the price of cook books and the sum of the price used in an expression:
```select type, price, price*2
from titles
where type like "%cook"
compute sum(price), sum(price*2)
```
```type price
------------ -------------- ------------
mod_cook 19.99 39.98
mod_cook 2.99 5.98
sum sum
============= ============
70.87 141.74
```
Usage
• The compute clause allows you to see the detail and summary rows in one set of results. You can calculate summary values for subgroups, and you can calculate more than one aggregate for the same group.
• compute can be used without by to generate grand totals, grand counts, and so on. order by is optional if you use the compute keyword without by. See Example 6.
• If you use compute by, you must also use an order by clause. The columns listed after compute by must be identical to or a subset of those listed after order by and must be in the same left-to-right order, start with the same expression, and not skip any expressions. For example, if the order by clause is``` order by a, b, c```, the compute by clause can be any (or all) of these:
```compute by a, b, c
compute by a, b
compute by a
```
Restrictions
• You cannot use more than 127 aggregate columns in a compute clause.
• You cannot use a compute clause in a cursor declaration.
• Summary values can be computed for both expressions and columns. Any expression or column that appears in the compute clause must appear in the select list.
• Aliases for column names are not allowed as arguments to the row aggregate in a compute clause, although they can be used in the select list, the order by clause, and the by clause of compute.
• You cannot use select into in the same statement as a compute clause, because statements that include compute do not generate normal tables.
• If a compute clause includes a group by clause:
• The compute clause cannot contain more than 255 aggregates
• The group by clause cannot contain more than 255 columns
• Columns included in a compute clause cannot be longer than 255 bytes.
compute results appear as a new row or rows
• The aggregate functions ordinarily produce a single value for all the selected rows in the table or for each group, and these summary values are shown as new columns. For example:
```select type, sum(price), sum(advance)
from titles
where type like "%cook"
group by type
```
```type
------------- --------- ----------
mod_cook 22.98 15,000.00
(2 rows affected)
```
• The compute clause makes it possible to retrieve detail and summary rows with one command. For example:
```select type, price, advance
from titles
where type like "%cook"
order by type
```
```type price advance
---------- ---------- ----------------
mod_cook 2.99 15,000.00
mod_cook 19.99 0.00
Compute Result:
---------------------- -----------------
22.98 15,000.00
---------- ---------- ----------------
Compute Result:
---------------------- -----------------
47.89 19,000.00
(7 rows affected)
```
• Table 7-4 lists the output and grouping of different types of compute clauses.
Table 7-4: compute by clauses and detail rows
Clauses and grouping
Output
Examples
One compute clause, same function
One detail row
1, 2, 4, 6, 7
One compute clause, different functions
One detail row per type of function
3
More than one compute clause, same grouping columns
One detail row per compute clause; detail rows together in the output
Same results as having one compute clause with different functions
More than one compute clause, different grouping columns
One detail row per compute clause; detail rows in different places, depending on the grouping
5
Case sensitivity
• If your server has a case-insensitive sort order installed, compute ignores the case of the data in the columns you specify. For example, given this data:
```select * from groupdemo
```
```lname amount
---------- ------------------
Smith 10.00
smith 5.00
SMITH 7.00
Levi 9.00
Lévi 20.00
```
compute by on lname produces these results:
``` select lname, amount from groupdemo
order by lname
compute sum(amount) by lname
```
``` lname amount
---------- ------------------------
Levi 9.00
Compute Result:
------------------------
9.00
lname amount
---------- ------------------------
Lévi 20.00
Compute Result:
------------------------
20.00
lname amount
---------- ------------------------
smith 5.00
SMITH 7.00
Smith 10.00
Compute Result:
------------------------
22.00
```
The same query on a case- and accent-insensitive server produces these results:
``` lname amount
---------- ------------------------
Levi 9.00
Lévi 20.00
Compute Result:
------------------------
29.00
lname amount
---------- ------------------------
smith 5.00
SMITH 7.00
Smith 10.00
Compute Result:
------------------------
22.00
```
Standards
SQL92 – Compliance level: Transact-SQL extension.
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http://www.jiskha.com/display.cgi?id=1290026595
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# consumer math
posted by on .
You are debating about whether to buy a new computer for \$800.00 or a refurbished computer with the same equipment for \$640.00. If a savings account earns 4.5% APR interest, how much do you really save with a refurbished computer if you put the difference into the savings account for a year?
• consumer math - ,
• consumer math - ,
\$800 - \$640 = \$160
So you invest \$160 for one year and get 4.5 % interest on it.
\$160 * 1.045 = \$167.20 in the bank after one years interst is added.
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http://mathoverflow.net/questions/116082/why-pullback-only-defined-up-to-isomorphism-but-nevertheless-presented-as-functor/116127
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## Why pullback only defined up-to-isomorphism but nevertheless presented as functor?
I am reading "Category Theory" (2nd ed.) of Awodey, and I'm stuck at page 96 (proposition 5.12) when pullbacks are presented as functors:
The pullback under question corresponds to this square:
$$\begin{matrix} C' \times_C A & \xrightarrow{h'} & A \\[1ex] \downarrow \rlap{\scriptstyle{\alpha'}} & & \downarrow\rlap{\scriptstyle{\alpha}} \\[1ex] C' & \xrightarrow{h} & C \end{matrix}$$
Here is the statement of Awodey's book that I do not understand:
Pullback is a functor. That is, for fixed $C' \rightarrow_h C$ in a category $\mathbf{C}$ with pullbacks, there is a functor
$h^* : \mathbf{C}/C \rightarrow \mathbf{C}/C'$
defined by
$(A\rightarrow_\alpha C) \mapsto (C'\times_C A \rightarrow_{\alpha'} C')$
where $\alpha'$ is the pullback of $\alpha$ along h
The problem that I see is that, given initially:
$$\begin{matrix} & & A \\[1ex] & & \downarrow \rlap{\scriptstyle{\alpha}} \\[1ex] C' & \xrightarrow{h} & C \end{matrix}$$
there can be several pullbacks on it, for example, in addition to $(\alpha',h')$, there could be $(\alpha_2',h_2')$, and the unique condition is that there exist an isomorphism i such that $\alpha_2' = \alpha\circ i$ and $h_2' = h'\circ i$. Worse, given a pullback $(\alpha',h')$, one can build as many as pullbacks as there exist isomorphisms, as given any isomorphism j (with domain $C' \times_C A \rightarrow_{h'}$) the two arrows $(\alpha'\circ j,h' \circ j)$ form a new pullback.
So, how could we build a functor if the image arrow is only defined up to an arbitrary isomorphism?
-
Just choose one pullback for each cospan, and show that any family of choices defines a functor. (That is why Awodey writes "_a_ functor"!) – Zhen Lin Dec 11 at 14:08
Zhen Lin: Yes, he writes a functor, but he also writes the pullback, so Almeo's point is well taken. The writing is indeed sloppy. Almeo: Zhen's solution does work, though: For each diagram, arbitrarily choose a pullback --- then you'll get a (non-uniquely-defined) functor. – Steven Landsburg Dec 11 at 14:16
No you won't get a functor, because functoriality will in general hold only up to isomorphism. – Andrej Bauer Dec 11 at 14:19
@Zhen: no, he write "a functor" because there are many functors, and pullback is one of them. Had he written "let $F$ be a pullback functor" then your comment would be relevant, as in this case we would be talking about one of many different pullback functors (all of which are naturlly isomorphic). I think it is not Awodey who is sloppy here. – Andrej Bauer Dec 11 at 14:22
Andrej: "No you won't get a functor"......you're right of course. – Steven Landsburg Dec 11 at 14:43
show 3 more comments
Let us first consider a slightly simpler situation. The cartesian product of sets $A$ and $B$ is a set $C$ with two maps $p_1 : C \to A$ and $p_2 : C \to B$ such that ... (familiar condition inserted here). All cartesian products of $A$ and $B$ are canonically isomorphic, and among them there is a particular one, denoted $A \times B$, which is specifically defined as $A \times B = \lbrace\lbrace\lbrace x,y \rbrace, \lbrace y\rbrace\rbrace \mid x \in A \land y \in B\rbrace$.
This is a familiar situation. Often a construction is determined up to canonical isomorphism, but we have a specific one that we can use as an operation, like $(A,B) \mapsto A \times B$ above.
Awodey does the same thing in his book. "Being a pullback" is a property, but we can turn it into structure, i.e., an operation which takes a pair of arrows $f : A \to C$ and $g : B \to C$ and gives a pullback square. You may wonder whether there always is such an operation. If you believe in the axiom of choice then the answer is positive, because we may always choose particular pullbacks among the canonically isomorphic ones. In concrete examples you will usually find chosen pullbacks easily, so this is not problematic.
There is a small catch. While for $h : A \to B$ it is the case that $h^{*}$ is a functor from $\mathcal{C}/B$ to $\mathcal{C}/A$, the assignment $h \mapsto h^\ast$ tends to be a "functor up to isomorphism" only. This is so because composition of chosen pullbacks need not be a chosen pullback (but is canonically isomprhic to it).
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Thank you very much for your answer !! Yes, you must be speaking about the functorial relation on compositions, aren't you? By the way, is that the relation that does not hold when we add fancy isomorphisms to all the $\alpha'$, as you replied to Zhen and Steven? So, any family of couples $(\alpha,\alpha')$ so that the square is always a pullback and so that the composition condition for functors is satisfied, is a definition of a functor for this pullback? – Almeo Maus Dec 11 at 15:00 Every answer brought me useful insight, so it is difficult to decide which is the "good one". However the answer of Andrej pointed that what I am stuck upon is the same notion that gives a product of two objects as an "operation" even when there are infinities of "product candidates", by taking one as a "reference" and calling it "canonical" then calling the others "canonically isomorphic" and the isomorphisms that link them to the canonical one "canonical isomorphisms". That does not answer all the pb but answers from Todd and David complete it. Moreover Andrej has published with Awodey :) – Almeo Maus Dec 13 at 6:30
Actually, it looks to me that there's a conflation of two different issues. According to the boxed statement in the OP, we just have to exhibit a functor $h^\ast: \mathbf{C}/C \to \mathbf{C}/C'$ for a fixed morphism $h: C' \to C$. We are not being asked to prove that we can choose a strict functor (as opposed to a pseudofunctor)
$$\mathbf{C}^{op} \to Cat$$
which takes each object $C$ to the slice $\mathbf{C}/C$, and morphisms $h$ to pullback functors, which appears to be the issue Andrej is discussing.
The issue being discussed in the boxed statement is easy, and can be boiled down to this: if $\mathbf{C}$ has pullbacks, then for each $h: C' \to C$ the pushforward functor $\sum_h: \mathbf{C}/C' \to \mathbf{C}/C$ (taking each object $f: X \to C'$ in the domain to the object $h \circ f: X \to C$ in the codomain, and defined in the obvious way on morphisms) has a right adjoint (which is of course a functor) $h^\ast$. Here we need only choose a pullback object $h^\ast g$ in $\mathbf{C}/C'$ for each object $g: Y \to C$ in $\mathbf{C}/C$, and then define $h^\ast$ on morphisms in the way dictated by the universal property. In other words, any choice of pullback $h^\ast g$, one for each object $g$ in $\mathbf{C}/C$, defines a universal arrow
$$\Phi_g: \sum_h (h^\ast g) \to g$$
so that having made these choices and given a morphism $f: g \to g'$ in $\mathbf{C}/C$ (i.e., a commutative triangle), we may then define $h^\ast f: h^\ast g \to h^\ast g'$ to be the unique arrow such that
$$\Phi_{g'} \circ \sum_h (h^\ast f) = f \circ \Phi_g$$
and functoriality of $h^\ast$ is assured by the usual universal arguments.
Edit: For example, let us show $h^\ast$ preserves compositions. Suppose given morphisms $f: g \to g'$ and $f': g' \to g''$ in $\mathbf{C}/C$. Then $h^\ast (f' \circ f)$ is the unique arrow $h^\ast g \to h^\ast g''$ such that
$$\Phi_{g''} \circ \sum_h h^\ast(f' \circ f) = f' \circ f \circ \Phi_g.$$
On the other hand,
$$\Phi_{g''} \circ \sum_h (h^\ast f' \circ h^\ast f) = \Phi_{g''} \circ (\sum_h h^\ast f') \circ (\sum_h h^\ast f) = f' \circ \Phi_{g'} \circ (\sum_h h^\ast f) = f' \circ f \circ \Phi_g$$
and so, by uniqueness, $h^\ast (f' \circ f) = (h^\ast f') \circ (h^\ast f)$.
-
Thank you very much for your answer! However, I have to take some time to understand it. The $h$ in the question was $h : C' \rightarrow C$, but is it intentionally that you took $h : C \rightarrow C'$ ? – Almeo Maus Dec 11 at 17:01 No, sorry, that was an accident in transcription. I can change it if you want (and hope there were no other errors in transcription). – Todd Trimble Dec 11 at 17:11 If it helps, the universality argument at the end is a special case of Categories for the Working Mathematician (2nd ed.), p. 83, Theorem 2 (iv). It's good to go through the argument at least once in one's life. – Todd Trimble Dec 11 at 17:17 Yes thank you, I think it would be more clear for me with the same notations as in the question (at least at the beginning). Thanks for the reference!! I will read it as soon as tomorrow. – Almeo Maus Dec 11 at 17:30 Done. I also spelled out the universal argument at the end. – Todd Trimble Dec 11 at 18:07
Why has no one mentioned anafunctors? They were invented by Makkai for the express purpose of expressing universal constructions as functor-like things (I was going to say 'objects') without having to make choices.
The definition is as follows: Let $C$ and $D$ be categories. An anafunctor from $C$ to $D$ is a span $C\leftarrow U \to D$ where $U \to C$ is fully faithful and surjective on objects.
That's it.
Of course, it is a bit more tricky to show that there are the arrows of a (weak) 2-category of categories, and that category theory works perfectly well using this notion of categories; for the present purposes, the references at the above link should satisfy the most curious.
In the case of the pullback 'functor' $h^*$ we have an anafunctor $\mathbf{C}/C \leftarrow P \to \mathbf{C}/C'$ where $P$ is the category with objects pullback squares with bottom arrow $h$ and morphisms the canonical thing which makes $P \to \mathbf{C}/C$ fully faithful: commutative triangles involving the other leg of the cospan involving $h$ - in other words, arrows in $\mathbf{C}/C$. This induces a canonical arrow in $\mathbf{C}/C'$ by the universal property of pullbacks, and then the functor $P\to \mathbf{C}/C'$ just forgets the original pullback square, and keeps the arrow with codomain $C'$. This is a functor by using the universal property of pullbacks a couple of times.
If one has a way of choosing a particular pullback square for each object of $\mathbf{C}/C$, say by some sort of canonical construction as in the category of (ZF-)sets (Kuratowski pairs and subsets), or by judicious amounts of the axiom of choice, then one can find a section on objects of $P \to \mathbf{C}/C$, and this gives you a canonical section of the functor $P \to \mathbf{C}/C$, which then gives a pullback functor $h^*\colon \mathbf{C}/C \to \mathbf{C}/C'$.
A similar sort of anafunctor exists for any universal construction, which is defined up to isomorphism by some universal property.
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@david: Has this idea been ap[plied to fibrations of categories replacing the notion of cleavage? I recall Benabou was not keen on assuming that fibrations had a cleavage. – Ronnie Brown Dec 12 at 11:32 Thank you very much for your answer! I will take time to understand the notion, that seems to help define those functors "defined up to (canonical) isomorphisms". – Almeo Maus 0 secs ago – Almeo Maus Dec 12 at 17:58 @Ronnie - there is definitely a relation to fibrations, but Jean Benabou would be furious if I didn't say that one can use what he calls distributors (but almost everyone else calls profunctors) to deal with this issue (and these have a longer history). There is a property of an anafunctor called saturation, and saturated anafunctors between categories are equivalent to representable distributors/profunctors, so you can approach it from either side. However, you don't need the relation between the two to recover the sort of thing you are thinking of using anafunctors. – David Roberts Dec 12 at 23:39 @Ronnie: It's true that every fibration admits a canonical "ana-cleavage". Is that what you were asking? – Mike Shulman Dec 15 at 20:41
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C4 Engine API Documentation
class Point3D
Defined in: `TSVector3D.h`
The `Point3D` class encapsulates a 3D point.
Definition
`class Point3D : public Vector3D`
Constructor
`Point3D();`
`Point3D(float a, float b, float c);`
Parameters
`a` The value of the x coordinate. `b` The value of the y coordinate. `c` The value of the z coordinate.
Description
The `Point3D` class is used to store a three-dimensional point having floating-point coordinates x, y, and z. The difference between a point and a vector is that a point is assumed to have a w coordinate of 1 whenever it needs to be converted to a four-dimensional representation, whereas a vector is assumed to have a w coordinate of 0. Such a conversion occurs when a vector or point is assigned to a `Vector4D` object or is multiplied by a `Transform4D` object.
The default constructor leaves the components of the vector undefined. If the values `a`, `b`, and `c` are supplied, then they are assigned to the x, y, and z coordinates of the base vector object, respectively.
The difference between two points produces a direction vector. A three-dimensional direction vector is converted to a point by adding it to the identifier `Zero3D`.
`Point3D& operator *=(float n);` Multiplies by the scalar `n`. `Point3D& operator /=(float n);` Divides by the scalar `n`.
`Point3D operator -(const Point3D& p);` Returns the negation of the point `p`. `Point3D operator +(const Point3D& p1, const Point3D& p2);` Returns the sum of the points `p1` and `p2`. `Point3D operator +(const Point3D& p, const Vector3D& v);` Returns the sum of the point `p` and the vector `v`. `Vector3D operator -(const Point3D& p1, const Point3D& p2);` Returns the difference of the points `p1` and `p2`. `Point3D operator -(const Point3D& p, const Vector3D& v);` Returns the difference of the point `p` and the vector `v`. `Point3D operator -(const Vector3D& v, const Point3D& p);` Returns the difference of the vector `v` and the point `p`. `Point3D operator *(const Point3D& p, float t);` Returns the product of the point `p` and the scalar `t`. `Point3D operator *(float t, const Point3D& p);` Returns the product of the point `p` and the scalar `t`. `Point3D operator /(const Point3D& p, float t);` Returns the product of the point `p` and the inverse of the scalar `t`. `Point3D operator *(const Point3D& p1, const Point3D& p2);` Returns the componentwise product of the points `p1` and `p2`. `Point3D operator *(const Point3D& p, const Vector3D& v);` Returns the componentwise product of the point `p` and the vector `v`. `Point3D operator *(const Vector3D& v, const Point3D& p);` Returns the componentwise product of the vector `v` and the point `p`.
`Vector3D` A `Point3D` object behaves much like a `Vector3D` object, but some properties are altered.
`Vector3D`
`Vector4D`
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You are on page 1of 5
# TestII PHY2020 Formula sheet
All problems worth 4 points total except #1; parts are worth the state
points!
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'
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'
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'
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'
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'
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'
7
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o
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o
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o
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3#2('C\$'!'#',,!' <! Qimilarl=> 400
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3 which is 2('C204!4 # 4((!4 <! Qo P2 # P1 14((!4&',,!' # 1!'0' P1
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4 Replies Latest reply: Sep 30, 2015 10:48 AM by
# What If Analysis - Simple query (I think)
Hi
I am sure this is a simple query for most, but I can't find the answer or any good qvw examples.
I have some data that tells me 'total hours worked' & 'total cost' over a period.
I want to create a slider that allows me allows me to change the number of hours, which then gives me a 'what if' on cost? i.e. How many hours less need to be worked to be within budget?
How do I do this?
Apologies if this is a really simple query, but I just can't find a good example.
Thanks
Phil
• ###### Re: What If Analysis - Simple query (I think)
Hi Phil,
Thinking through the logic, you'd want to create a variable (call it vHours) and create a slider to manage this variable. You can either enter logical minimum/maximum values for the slider, or create a range based on your data...something like Sum([Total Hours Worked]) * X where X is your stretch factor.
I assume you'd want a "cost per hour", so create an expression either in a text box (if this is aggregated data) or as an expression in a straight table chart (where project or person is your dimension for example) that is [total cost]/[total hours worked].
If you then multiply this by your variable you'll see the new "total" cost change depending on the number of hours worked. You can then match it to whatever budgets you have.
Happy to mock this up if that doesn't make sense!
• ###### Re: What If Analysis - Simple query (I think)
Plzz help me... if any1 know about the what-if-queries... how does it work in a query...
• ###### Re: What If Analysis - Simple query (I think)
Help me!! If any1 knows about whatif query works in a given query.. why do you used it???
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How much mechanical energy did lose on the hill
A fun-loving 11.4 kg otter slides up a hill and then back down to the same place. she starts up at 5.75m/s and returns at 3.75m/s , how much mechanical energy did she lose on the hill? change of (E (mech))= ? J what happened to the energy?
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Every one of the support staff and the participants have to
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Every one of the support staff and the participants have to compulsorily go through a full body search before they enter the competition area.
a. have to compulsorily go through a full body search before they enter
b. have to go through a compulsorily full body search before entering
c. have to compulsorily go through a full body search before he or she enters
d. has to compulsorily go through a full body search before entering
e. has to compulsorily go through a full body search before they enter
Please explain and give other examples to understand this.
I do know that 'every one' is singular but with a conjunction followed by a plural noun
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11 Jan 2013, 00:09
Each/Every changes the verb into singular (i.e. overrides the number rule for conjunction)
For example: Every boy and girl is required to wear an uniform.
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11 Jan 2013, 02:39
I believe there is some ambiguity in the question statement itself. If "every one of the support staff" and "the participants" are different entities, "have" could well be the right usage.
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Re: every one [#permalink] 11 Jan 2013, 02:39
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# Trying to return value from another sheet with ranges
May 17, 2012 at 19:29:37
Specs: Windows 7
Could someone please help me!I am trying to figure out what I'm doing wrong. I want to have the user enter a loan amount and a percentage, then in my spreadsheet i have the particular rate appear that corresponds to information inputed above. =IF(AND(C2<=250000),IF(AND(F2<=75%),DATA!J4),IF(OR(C2<=250000),IF(AND(F2>=76%,F2<=90%),DATA!J5,DATA!J6)))This is what I have so far.If loan amount is under 250,000 and greater than or equal to 76% and less than equal to 90% then return a figure on the sheet named Data J5. I have other conditions that I have to add to this as well. However if I can work out how to do these, then I should be ok.Can anyone help me please
See More: Trying to return value from another sheet with ranges
#1
May 17, 2012 at 19:48:01
The syntax for your AND and OR is wrong.Read the Excel Help files to see how they should be used.Since AND and OR are used to compare 2 or more objects, using something like:AND(C2<=250000)makes no sense.Perhaps you are looking for something like this:IF(AND(C2<=250000, F2<=75%)....which means that both of those conditions must be TRUE for the IF to be TRUE.For OR, it would be:IF(OR(C2<=250000, F2<=75%)....which would mean that any one of those would have to be TRUE for the IF to be TRUE.Click Here Before Posting Data or VBA Code ---> How To Post Data or Code.
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#2
May 17, 2012 at 19:55:01
Thanks for the help, I tried the following:-=IF(AND(C2<=250000),IF(AND(F2<=75%),DATA!J4),IF(AND(C2<=250000,F2>=76%,F2<=90%),DATA!J5,))It is still resulting in a false statement in the cell, when it should be producing a rate.Are you able to shed any light on this? Thanks
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Diophantine Pairs Once upon a time, there was a chap called Diophantus (250 A.D.), who was interested in the problem of solving equations which had a whole range of possible answers. It sounds as if this would make the Problem easier, not harder, but the real Problem wasn’t finding an answer, but deciding how many answers there were altogether. We now use his name to describe any equations whose (multiple) solutions have to be Integers. The Problem: Solve these Diophantine equations, where x and y must be positive Integers. 1. (Not many answers…) 2. (Lots of answers e.g. x = 48, y = 9 - find as many as you can) Open the File as a Word Document
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# Data validation allow weekday only
We can use the Excel WEEKDAY function to ensure that data validation allows weekday only. The steps below will walk through the process of using Data Validation to allow weekday only.
Figure 1 – Example of Data Validation to allow weekday only
## General Formula
`=WEEKDAY(A1,2)<6`
## Formula
`=WEEKDAY(C5,2)<6`
## Setting up the Data
• We will set up our data in Columns B and C
• Column B will contain our Groceries
• Column C will contain our dates of purchase
Figure 2 – Setting up the Data
## Applying the WEEKDAY customized function
• We will highlight Cells C5 to C7
• We will click on Data Validation in the Data tab within the ribbon at the top of the worksheet
Figure 3 – Clicking on Data tab
• We will check the ignore blank box, select custom, and enter the formula below in the formula box
`=WEEKDAY(C5,2)<6`
• We will press OK
• This will allow weekdays only
Figure 4 – Applying the WEEKDAY customized function
## An Alternative Formula for Data validation
We can also create a return value argument that would allow the dates from 1 (Sunday) to 7 (Saturday).
• We will again highlight on Cells C5 to C7 and select Data Validation
• We will select custom, and enter this formula below in the formula dialog box
`=AND(WEEKDAY(B2)<>1,WEEKDAY(B2)<>7)`
• We will press OK. This would return TRUE for all values if they fall within Saturday to Sunday.
Figure 5 – Alternative formula for Data Validation
## Explanation
We use the Data validation rules to restrict our cells so any user can only add values within the specific restriction. For our formula, the return value, 2 is a Monday-based number. This would be compared by Excel against 6. Therefore, any values less than 6 will return true. If the value is more than 6, the date is Sunday or Saturday, then validation fails.
## Note
• If we wish to allow only dates that occur in the weekend such as Sunday or Saturday, then we can use this formula:
`=WEEKDAY(C5,2)>5`
• The Cell reference in the data validation formulas are always relative to the upper left cell, hence the reason our formula contains, C5.
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RRB ALP & Technician Mock Test (English) - 1 - Railways MCQ
# RRB ALP & Technician Mock Test (English) - 1 - Railways MCQ
Test Description
## 30 Questions MCQ Test - RRB ALP & Technician Mock Test (English) - 1
RRB ALP & Technician Mock Test (English) - 1 for Railways 2024 is part of Railways preparation. The RRB ALP & Technician Mock Test (English) - 1 questions and answers have been prepared according to the Railways exam syllabus.The RRB ALP & Technician Mock Test (English) - 1 MCQs are made for Railways 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB ALP & Technician Mock Test (English) - 1 below.
Solutions of RRB ALP & Technician Mock Test (English) - 1 questions in English are available as part of our course for Railways & RRB ALP & Technician Mock Test (English) - 1 solutions in Hindi for Railways course. Download more important topics, notes, lectures and mock test series for Railways Exam by signing up for free. Attempt RRB ALP & Technician Mock Test (English) - 1 | 75 questions in 60 minutes | Mock test for Railways preparation | Free important questions MCQ to study for Railways Exam | Download free PDF with solutions
RRB ALP & Technician Mock Test (English) - 1 - Question 1
### Which one of the following is acidic?
RRB ALP & Technician Mock Test (English) - 1 - Question 2
### Six years ago, Seema was half of that of Rupa in age. Four years hence the respective ratio of their ages would be 3:5. How old is Rupa at present?
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 2
Let Seema's age at present be x years and Rupa's age be y years,
Now, according to the question,
(y - 6) = 2 (x - 6)
⇒ y - 6 = 2x - 12
or, 2x - y = 6 .....(1)
Also, x + 4/y + 4 = 3/5
or, 5x + 20 = 3y + 12
or, 5x - 3y = -8 .....(2)
From Equations (1) and (2), we get
x = 26 years and y = 46 years
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RRB ALP & Technician Mock Test (English) - 1 - Question 3
### Enzyme which converts starch into glucose is:
RRB ALP & Technician Mock Test (English) - 1 - Question 4
Select the related word/figure from the given alternatives.
Lion : Den : : Rabbit :?
RRB ALP & Technician Mock Test (English) - 1 - Question 5
Most fish do not sink in water because of the presence of (1) swim bladder (2) air bladder (3) air sacs (4) air in spongy bones
RRB ALP & Technician Mock Test (English) - 1 - Question 6
Pointing to a man, a woman said : " His mother is the only daughter of my mother ." How is the woman related to the man?
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 6
Only daughter of my mother → myself.
Hence, the woman is the mother of the man.
RRB ALP & Technician Mock Test (English) - 1 - Question 7
Area of a square field is 1 hectare. The length of another square field is 1% more than the length of this square field. The difference in their areas is
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 7
1 hectare = 10000 m2
Area of a square field = 10000 m2
let length of a square field be x m
then x2 = 10000
x = 100m
length of other square field = 100 + 100 x 1/100 = 101
Area of Another square field = (101)2 = 10201 m2
∴ Difference in their areas = 10201 - 10000
= 201 m2
RRB ALP & Technician Mock Test (English) - 1 - Question 8
On what dates of April, 1995 did Sunday fall?
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 8
1993 complete years = 1600 + 300 + 93
= 1600 + 300 + 23 leap years + 70 ordinary yrs.
= (0 + 1 + 23 x 2 + 70) odd days
= 117 odd days
= 16 weeks + 5 odd days
Hence Jan 1, 1994 was a Friday, Jan, February, March have 3, 0 and 3 odd days respectively, or 6 odd days.
Hence, April 1, is a Thursday. April 4 is a sunday Option (c) is correct.
RRB ALP & Technician Mock Test (English) - 1 - Question 9
In which of the following cell organelles do photo and thermochemical reactions occur in different sites?
RRB ALP & Technician Mock Test (English) - 1 - Question 10
Which of these will cause a chemical change to occur?
RRB ALP & Technician Mock Test (English) - 1 - Question 11
Choose the word that does NOT match with the two given words.
Querulous : Peevish
RRB ALP & Technician Mock Test (English) - 1 - Question 12
Choose the numeral pair which is different from others .
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 12
Except 9 - 3, in all other pairs, the first number is four times the second number.
RRB ALP & Technician Mock Test (English) - 1 - Question 13
If 'green' is called 'white' , 'white' is called 'yellow', 'yellow' is called 'red', 'red' is called 'orange', then which of the following represents the colour of sunflower?
RRB ALP & Technician Mock Test (English) - 1 - Question 14
In a certain code, COMPUTER is written as RFUVQNPC. How is MEDICINE written in same code ?
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 14
RRB ALP & Technician Mock Test (English) - 1 - Question 15
Complete the analogous pair.
Forecast : Future : : Regret : ?
RRB ALP & Technician Mock Test (English) - 1 - Question 16
The difference between compounded interest and simple interest on a certain sum for one year at 5% per 6 months is Rs 3. The sum is
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 16
RRB ALP & Technician Mock Test (English) - 1 - Question 17
Local self-government can be best explained as an exercise in
RRB ALP & Technician Mock Test (English) - 1 - Question 18
Nayachar is an emerged island in
RRB ALP & Technician Mock Test (English) - 1 - Question 19
Swayam Prabha’, a Government of India initiative, is a project for
RRB ALP & Technician Mock Test (English) - 1 - Question 20
Which among the following financial services player has recently received the certificate of registration from the National Housing Bank to carry out housing finance activities?
RRB ALP & Technician Mock Test (English) - 1 - Question 21
The 21st World Congress of Mental Health was organized in which city?
RRB ALP & Technician Mock Test (English) - 1 - Question 22
Name the eminent scientist, well-known as Father of DNA finger printing in India, who passed away recently due to health problem
RRB ALP & Technician Mock Test (English) - 1 - Question 23
In a certain code language, '297' means 'tie clip button'. Which number means 'button' ?
I. In that language '926' means 'clip your tie'.
II. In that language '175' means 'hole and button'.
The question given above has a problem and two statements giving certain information. You have to decide if the information given in the statements is sufficient for answering the problem. Indicate your answer as
RRB ALP & Technician Mock Test (English) - 1 - Question 24
A and B starts moving towards each other from two places 200 m apart. After walking 60 m, B turns left and goes 20 m, then he turns right and goes 40 m. He then turns right again and comes back to the road on which he had started walking. If A and B walk with the same speed, what is the distance between A and B now ?
RRB ALP & Technician Mock Test (English) - 1 - Question 25
Based on the latest State Bank of India (SBI) Ecowrap report, the GVA estimate for current fiscal is estimated to trend higher at _____ percent from the projected 6.1 per cent
RRB ALP & Technician Mock Test (English) - 1 - Question 26
Usually, the colours used for live, neutral and earth wire respectively are
RRB ALP & Technician Mock Test (English) - 1 - Question 27
If 2x4 + 9 x3 + 8 x2 + 9 x + 2 = 0 ; then x =
Detailed Solution for RRB ALP & Technician Mock Test (English) - 1 - Question 27
Divide throughout by x2
RRB ALP & Technician Mock Test (English) - 1 - Question 28
The expression sin 89o + tan 89o , when expressed in terms of angles between 0o and 45o, can be written as
RRB ALP & Technician Mock Test (English) - 1 - Question 29
Value of the following is
RRB ALP & Technician Mock Test (English) - 1 - Question 30
Which fruit can diabetic patient eat freely?
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# systems of equations word problems worksheet answer key with work
``` . Word Problems Using Systems of Equations 34. Children’s tickets cost y dollars. Solving Systems of Equations Real World Problems. Simultaneous Equations Word Problems Worksheet With Answers Valid 40 from solving systems of equations by elimination worksheet answers with work , source:curiousmind.co . Solving Multi Step Equations Word Problems Worksheet Answers Once you do that, these linear systems are solvable just like other linear systems. Some of the worksheets for this concept are Systems word problems, Word problem practice workbook, Using equal values method or substitution to find the, , Systems of equations by substitution, Practice solving systems of equations 3 different, Model practice challenge problems vi, Grades mmaise salt lake city. The worksheets on this page have four coordinate planes and systems equations in point-slope form that students graph to solve, and includes an answer key showing the correct graph. 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# Recent content by las3rjock
1. ### Guess I'm going to have to give in and learn a language
Since you're starting off with no programming knowledge, I would probably suggest a book like Mark Lutz's Learning Python. I learned Python using Alex Martelli's Python in a Nutshell and David Beazley's Python Essential Reference, but I already had programming experience in C, Java, and MATLAB.
2. ### Help which language best suits my situation?
Based on your rather vague description of your requirements, MATLAB probably is the best fit: http://en.wikipedia.org/wiki/MATLAB
3. ### LaTeX Latex table help
http://en.wikibooks.org/wiki/LaTeX/Tables#Rows_spanning_multiple_columns
4. ### Optics, image
This part of the problem statement means that the object is located 1.2 m in front of the first refracting surface of the crystal ball (1.3 m from the center of the crystal ball, and 1.4 m from the second refracting surface of the crystal ball). The displacements z0 and z1 in the equation are...
5. ### Optics, image
To get you started, the equation for the first surface is \frac{1.5}{z_1} - \frac{1}{-1.2} = \frac{1.5 - 1}{0.1} which you should solve for z1. This gives a location for the image formed by the first surface relative to the vertex of the first surface. Then you treat that image as the object...
6. ### Optics, image
Born & Wolf is way too advanced a book to be using to solve a problem like this, but the equation you want is in section 4.4.1 (in the sixth edition--it may be different in other editions), "Refracting surface of revolution," and it's the one that looks like this \frac{n_1}{z_1}-\frac{n_0}{z_0}...
7. ### LaTeX LaTeX sometime the arrow is over the right
$g(\vec{t})$ Note the curly braces.
8. ### MikTex 2.8 for beginners
MikTeX now comes with a basic LaTeX document editor called TeXworks, which you should find under "Start Menu"->"All Programs"->"MikTeX 2.8". Run it, and enter the following document: \documentclass{article} \begin{document} \title{Example Document} \author{John Doe} \maketitle This is...
9. ### A four bit adder logic circuit block diagram
Alternatively, a truth table for the sum s and the carry bit co for the addition of two 1-bit numbers a and b could be illuminating... :wink:
10. ### LaTeX LaTex for windows?
http://wiki.lyx.org/Windows/Windows
11. ### Steam Tables For Apple Computers
http://freesteam.sourceforge.net/
12. ### MATLAB Create a Hologram with Matlab?
It is not the superposition of the fields (the addition process O+R) that results in interference, but the determination of the intensity of the superimposed fields (finding the modulus squared of the total field \lvert O+R \rvert^2). Let O(x,y) = o(x,y) \exp [i \phi(x,y)] where o(x,y) and...
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37s
## The golden ratio spinning top
This is the φ (phi) top. It's made of solid brass in the shape of a prolate elIipsoid- and the name of the top is derived from the following design requirement: the ratio of the semi minor (short) axis to the semi major (long) axis is made equal to the mathematical constant the golden ratio φ. Invented by astrophysicist Kenneth Brecher, this top stands up vertically (when spun with sufficient rotational velocity) due to physics similar to that of the tippe-top. The concave mirror keeps the top from wandering off and the glass surface has minimal friction allowing the top to spin for a long time- this video shows various stages of 8 minutes of total spin time.
37s
## Crazy experiment shows what a magnetic field looks like
This is a perfect and simple visual representation of a magnetic field. Iron filings are suspended in oil, and you can see them gracefully line up along the magnetic field lines when the ferromagnet is dropped into the setup, revealing a clear dipole configuration.
23s
## Watch as this crazy spinning top starts levitating
The Levitron: spin stabilized magnetic levitation- will fly for about 3 minutes or more per launch. The spin rate of the magnetic top must be just right for flight, not too fast or too slow. Also the weight of the top must be adjusted to within 1 gram to achieve equilibrium between the pull of gravity and the magnetic repulsion from the large donut shaped magnet in the wood base. Probably my favorite physics toy of all time. Find out more: http://physicstoys.blogspot.com/p/physics-toy-store.html
16s
## Physics Fun: Balance Pteranodon
Weights in the forward wing tips create a center of mass below this pterosaur's beak- making a system that can rotate and oscillate about a stable equilibrium condition. No magnets used or needed for this physics! Usually this toy is found in the form of a bird (direct dinosaur descendant). Pterosaurs are not dinosaurs, but are the largest animals to ever fly, and powered flight has only evolved four times: bats, birds, pterosaurs, and insects- in reverse order. Follow @PhysicsFun for more.
37s
## The Strange Physics of the φ Top
This unique top is made of solid brass in the shape of a prolate elIipsoid- and the name of the top is derived from the following design requirement: the ratio of the semi minor (short) axis to the semi major (long) axis is made equal to the mathematical constant the golden ratio φ (phi). Invented by astrophysicist Kenneth Brecher, this top stands up vertically (when spun with sufficient rotational velocity) due to physics similar to that of the tippe-top. The concave mirror keeps the top from wandering off and the glass surface has minimal friction allowing the top to spin for a long time- this video shows various stages of 8 minutes of total spin time. Follow @PhysicsFun for more.
49s
## Plant Phone Charger
This pot plant can charge your phone through photosynthesis.
43s
## This fish tank is filled with ingenious 3D-printed coral
Whether your lease doesn't allow pets, or your trying to convince your kid that cats are really hard to take care of, a pet fish is the way to go.
37s
## Airbus has just created a jet that runs on algae
A new plane by Airbus runs on fuel exclusively produced by algae.
49s
## Plant communication
Scientists have invented a machine that lets us communicate with plants.
47s
## Tiny Drone
Scientists have invented a tiny flying robot that can perch to save energy.
41s
## Flaming Crater
For more than 40 years, this natural gas crater in Turkmenistan has been constantly burning - and scientists aren't really sure why.
52s
## Growing Bricks
This company is growing bricks that fight climate change
43s
## Solar Bottle Lamps
These simple water bottle lamps are bringing light to those in need.
54s
## Lunar Globe
This topographically accurate 'lunar globe' is the coolest way to explore the Moon.
54s
## Chemical 3D Printer
This company is growing drones in giant chemical vats
45s
## Plastic Eating Mushrooms
These mushrooms eat plastic waste - and then you can eat them.
43s
Turns out dung beetles navigate using the stars.
49s
## Weird Liquids in Space
Watch astronauts have the best time ever with liquid
41s
## Glass Igloo
This glass igloo hotel lets you watch the Northen Lights in bed.
47s
## Gravitational Waves
This is how scientists detected gravitational waves for the second time.
1m27s
A new patent means Google Glass might soon have night vision.
48s
## Carbon to Stone
Here's how engineers are turning CO2 into stone.
55s
## Cholesterol Machine
This Machine Could Remove Plaque From Your Arteries in Minutes.
50s
## Floating Home
This floating home lets you travel across the world off-grid.
45s
## Lockheed Martin SchoolBus
This school bus takes kids on a virtual trip to Mars.
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# Reduce 6058
If we reduce the number x by 30%, we get 140. What is x?
x = 200
### Step-by-step explanation:
x-30/100 x = 140
x-30/100·x = 140
70x = 14000
x = 14000/70 = 200
x = 200
Our simple equation calculator calculates it.
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips for related online calculators
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function [wh,wo,tstcol]=doclassif(trndat,trnlab,tstdat,tstlab,U,I,E,EL,R) % [H,O,TO] = doclassif(D,L,TD,TL,U,I,E,EL,R) Try classification % D is rows of training data, with labels given by L. % TD is rows of test data, with oracle results given by TL. % Train a neural net on D with E epochs of I iterations, % starting with learning rate = R, then halving it each epoch after % the first EL. Use U hidden units. % Returns the neural net in H and O, and the test output classes in TO % 2001-03-27 dpwe@ee.columbia.edu if nargin < 5 U = 5; end if nargin < 6 I = 29; end if nargin < 7 E = 4; end if nargin < 8 EL = 1; end if nargin < 9 R = 0.08; end % Make actual target matrix [ntrn,ndim] = size(trndat); trg = sparse([1:ntrn],trnlab,1); % Calculate norms nrm = [mean(trndat);std(trndat)]; % Do training wh = U; wo = []; for ep = 1:E if ep > EL R = R / 2; end [wh,wo] = nntrain(trndat,nrm,trg,R,I,wh,wo,16,10); % Do testing % Measure accuracy on train data to start trnop = nnfwd(trndat,nrm,wh,wo); trncol = findmaxcol(trnop); tfac = mean(trncol == trnlab); % Now do real test on test data % Calculate net output for test data (using norms from trn data) tstop = nnfwd(tstdat,nrm,wh,wo); plot(tstop); % Find 'winning' output for each pattern tstcol = findmaxcol(tstop); % Calculate proportion of matches to oracle labels efac = mean(tstcol == tstlab); disp(['Ep ',num2str(ep), ' lr=', num2str(R), ' frame accuracy on trn = ', num2str(100*tfac),'% test = ', num2str(100*efac),'%']); end
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## 17 October 2007
### Extraordinary claims require extraordinary evidence
Here's a book that's available online that I learned about recently: Information Theory, Inference, and Learning Algorithms by David MacKay. I learned about it from Steve at Information Processing; he learned about it from Nerd Wisdom.
I rather like the trend of books being available online. For one thing, it means that I do not have to carry books back and forth between my office and my apartment. (This is unfortunately not true for the calculus book I'm teaching from, which is about 1200 pages; fortunately my officemate has a copy of the book he doesn't use, so I use his copy when I'm on campus and my copy when I'm at home.)
This book seems to be a good introduction to information theory, machine learning, Bayesian inference, etc.; I have not had the chance to read any part of it thoroughly but I have randomly sampled from it and it seems quite interesting.
A few days ago I wrote about the question of finding the next element in the sequence "1, 2, 5, ..." and some of my readers complained that this problem is not well-posed. MacKay, in his chapter on Occam's razor, gives a similar example: what's the next number in the sequence "-1, 3, 7, 11"? You probably say 15. Consider two possible theories -- the sequence is an arithmetic sequence, or it is given by a cubic function of the form cx3 + dx2 + e, where c, d, and e are rational numbers. (The omission of the linear term is deliberate; otherwise the first theory would be a case of the second one.) The first one turns out to be much more likely, for the simple reason that there are less parameters to be tweaked! Let's say it's equally likely that the underlying function is linear or cubic; there are just a lot more possible cubic functions, so each particular cubic function is less likely. (For the details, see p. 345 of MacKay.)
By logic such as this, the most likely next element in that sequence is difficult to say, though... should we prefer 10, because then the nth term is given by the simple explicit formula (n-1)2+1? Or should we prefer 12 or 13, which are both given by simple linear recurrences? My instinct is that it depends on where the problem comes from, since the various possible next terms arise from different sorts of combinatorial structures, and in this sense the problem was ill-posed. In reality we wouldn't start out by assuming that all possible theories have equal probability; for one thing, there are infinitely many of them! The "simple" theories (the sequence has an explicit formula which is a polynomial, or a linear recursion with constant coefficients, or something like that) have higher prior probabilities... but given enough evidence, any complicated theory that starts out with nonzero probability of being true could turn out to be true. Extraordinary claims, though -- those with low prior probability -- require extraordinary evidence. There's a nice toy example a little further on in MacKay (p. 351) showing that if you see a piece of a box sticking out to the left of a tree, and a piece of a box of the same height and color sticking out to the right, it is nearly certain that those are actually pieces of the same box, and not two different boxes.
(What I've sketched in the previous paragraph is a bit of a lie, though; mathematical reasoning is rarely anywhere near this fuzzy. One could almost argue that it never is, because if we are so unsure about whether a result is true or not then we just don't call it mathematics.)
PS: I realized that I had a bunch of things I wanted to write about that were kind of piling up, so I might be digging into stuff that made its way through the blogosphere a month or so ago. I would feel no need to apologize for this except that blogs move so quickly.
PPS: The title of this post is due to Carl Sagan. I did not realize that when I titled the post, though, but I knew I'd heard the phrase somewhere before and Google tells me it's his.
#### 1 comment:
John Armstrong said...
The first one turns out to be much more likely, for the simple reason that there are less parameters to be tweaked!
Well, "likely" here is an uncharacteristic lapse of your usual (and appreciated!) pedantic use of probability and statistics terms.
Anyhow, so you're saying that Occam's razor is the opposite of the second law of thermodynamics?
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# Multi step word problems 3rd grade ideas
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This Is A Set Of Worksheets For The Third Grade Math Standard 3 Oa D 8 Which Is Solve Two Step Word P Word Problems 3rd Grade Word Problems Math Word Problems From pinterest.com
Multistep Word Problems 3rd Grade - Displaying top 8 worksheets found for this concept. I always find that providing a seasonal worksheet helps keep my daughter excited about doing Worksheets work. Solving Multi-Step Word Problems. Click on the images to view download or print them. Interpret remainders and their effect on the quotient. Third Grade Word Problems Worksheets and Printables.
### Social studies science and the Olympics are just some of the themes that will stimulate third graders as they apply addition subtraction and multiplication.
Decide what operation to use. As students complete one problem I provide other problems for them to complete. This collection of worksheets will help kids grasp how math applies in real world situations. Using different problems for each student or pair of students allows them to work independently have math conversations and discuss different types of solutions. Students will need to add and subtract 2-digit and 3-digit numbers to find the answers to these multi-step word problems. Multistep Word Problems 3rd Grade.
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Multi-Step Word Problems - 3rd Grade All Four Operations This packet contains 24 multi-step word problems where students have to add subtract multiply and divide numbers. Page 3 of 45 GreatSchools Students who are preparing for a Common Core Third Grade Mathematics exam should have a working knowledge of four important components. 3rd Grade Multi-Step Word Problems Intermediate Level Mom shares a packet of 18 cookies equally among her 3 daughters Emma Mia and Flora. As students complete one problem I provide other problems for them to complete. 2nd through 4th Grades.
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Multiple Step Basic 2. Multi-step word problems are Math of the greatest challenges 3rd many students. Multistep Word Problems 3rd Grade - Displaying top 8 worksheets found for this concept. If youre seeing this message it means were having trouble loading external resources on our website. In this fourth-grade learning game Jumpy.
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Each word problem includes a series of problem solving steps. Sample word problems can be found on the K-5 Math Teaching Resources website. Math word problems are a great way to help learners apply their math skills to real-world scenarios. Multi-Step Word Problems players get lots of practice solving mixed operation word problems in their quest to help Roly make it to the party waiting for him on the other side of the obstacles. 2nd through 4th Grades.
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2nd and 3rd Grades. Multi-step word problems are Math of the greatest challenges 3rd many students. Keywords can help you figure out which operation to use. This collection of worksheets will help kids grasp how math applies in real world situations. Here is a list of some of the common keywords used in word problems.
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Word problems for grades 3rd - 5th that take more than one step to solve. Using different problems for each student or pair of students allows them to work independently have math conversations and discuss different types of solutions. Third Grade Word Problems Worksheets and Printables. A Problems description of the worksheets is on each of the worksheet widgets. Click on the images to view download or print them.
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Word problems for grades 3rd - 5th that take more than one step to solve. 3rd Grade Math Story Problems - 3rd grade Worksheets word lists and activities. Decide what operation to use. Here is a list of some of the common keywords used in word problems. Word Problems Algebra - 3rd Grade Math Word Problems.
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Third Grade Word Problems Worksheets and Printables. These word problems are designed to help 3rd grade students exercise multi-step math Probblems solving reasoning critical thinking and precise math Grade skills. I always find that providing a seasonal worksheet helps keep my daughter excited about doing Worksheets work. Solving Multi-Step Word Problems. Sample word problems can be found on the K-5 Math Teaching Resources website.
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Word Problems Algebra - 3rd Grade Math Word Problems. Solve multi-step word problems using any of the four basic operations. Multi-step word problems are Math of the greatest challenges 3rd many students. Some of the worksheets for this concept are Multiple step problems Multiple step problems Word problems work easy multi step word problems Two step word problems Word problem practice workbook Fractionwordproblems Grade 3 math. Multi-Step Word Problems - 3rd Grade All Four Operations This packet contains 24 multi-step word problems where students have to add subtract multiply and divide numbers.
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Multi Step Math Word Problems - Fraction Word Problems 6th Grade Word problems allow students the opportunity to apply their math skills in authentic situations. Third Grade Word Problems Worksheets and Printables. Multi Step Addition And Subtraction Word Problems Y5 Worksheets - there are 8 printable worksheets for this topic. This collection of worksheets will help kids grasp how math applies in real world situations. Social studies science and the Olympics are just some of the themes that will stimulate third graders as they apply addition subtraction and multiplication.
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Worksheets are Y5 multi step Addit. Here is a list of some of the common keywords used in word problems. Worksheets are Y5 multi step Addit. All too often children who are able to solve numeric problems find themselves at a loss when faced with a word problem. Interpret remainders and their effect on the quotient.
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I always find that providing a seasonal worksheet helps keep my daughter excited about doing Worksheets work. Solving word problems can be fun and challenging They are like real-life problems. In this fourth-grade learning game Jumpy. Multistep Word Problems 3rd Grade. These steps were created for the purpose of making sense of.
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Multiple Step Basic 2. These multi-step questions can be calculated using only addition and subtraction. Solving word problems can be fun and challenging They are like real-life problems. If youre seeing this message it means were having trouble loading external resources on our website. Multi-step word problems are Math of the greatest challenges 3rd many students.
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3rd Grade Math Word Problems Worksheets - Math Problem Solving for 3rd Grade. Multi Step Math Word Problems - Fraction Word Problems 6th Grade Word problems allow students the opportunity to apply their math skills in authentic situations. Multiple Step Word Problems 3rd Grade - Displaying top 8 worksheets found for this concept. If youre seeing this message it means were having trouble loading external resources on our website. Whenever solving word problems you have to.
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Mass Volume and Capacity activity packet is designed to provide students in Grade 5 with hands-on. Flora eats 2 cookies right away. Solving Multi-Step Word Problems. 2nd through 4th Grades. Multi-Step Word Problems players get lots of practice solving mixed operation word problems in their quest to help Roly make it to the party waiting for him on the other side of the obstacles.
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Solve multi-step word problems using any of the four basic operations. Students will need to add and subtract 2-digit and 3-digit numbers to find the answers to these multi-step word problems. Solve multi-step word problems using any of the four basic operations. 3rd Grade Math Story Problems - 3rd grade Worksheets word lists and activities. Word problems for grades 3rd - 5th that take more than one step to solve.
Source: pinterest.com
Here is a list of some of the common keywords used in word problems. Solve multi-step word problems using any of the four basic operations. Multistep Word Problems 3rd Grade. These word problems are designed to help 3rd grade students exercise multi-step math Probblems solving reasoning critical thinking and precise math Grade skills. Basic multiplication and division fractions finding the 3rd of a rectangle 3rd familiarity with a variety of angles and geometric.
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Some of the worksheets for this concept are Multiple step problems Multiple step problems Word problems work easy multi step word problems Two step word problems Word problem practice workbook Fractionwordproblems Grade 3 math. Page 3 of 45 GreatSchools Students who are preparing for a Common Core Third Grade Mathematics exam should have a working knowledge of four important components. Multistep Word Problems 3rd Grade - Displaying top 8 worksheets found for this concept. Multi-step word problems are Math of the greatest challenges 3rd many students. Word Problems Algebra - 3rd Grade Math Word Problems.
Source: pinterest.com
Decide what operation to use. 3rd Grade Math Story Problems - 3rd grade Worksheets word lists and activities. 3rd Grade Math Word Problems Worksheets - Math Problem Solving for 3rd Grade. All too often children who are able to solve numeric problems find themselves at a loss when faced with a word problem. Multi-Step Word Problems - 3rd Grade All Four Operations This packet contains 24 multi-step word problems where students have to add subtract multiply and divide numbers.
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# Tutorial :Is it a good idea to use IEEE754 floating point NaN for values which are not set?
### Question:
Is it a good idea to use IEEE754 floating point NaN (not-a-number) for values which are undefined for non-mathematical reasons?
In our case they are not yet set because the values have not been received from some other device. The context is an embedded system using IEC1131 REAL32 values. Edit: The programming language is C, so we would most likely use NAN and isnanf(x), which are from C99. Though we may need some extra contortions to get these into our OS compatibility layer.
The default in programming languages seems to be to initialize floating point variables with positive zero, whose internal representation is all zeros. That is not usable for us, because 0 is in the range of valid values.
It seems like a clean solution to use NaN, but maybe it is more hassle than it is worth and we should pick some other value?
### Solution:1
Just noticed this question.
This is one of the uses of NaNs that the IEEE 754 committee has in mind (I was a committee member). The propagation rules for NaNs in arithmetic make this very attractive, because if you have a result from a long sequence of calculations that involve some initialized data, you will not mistake the result for a valid result. It can also make tracing back through your calculations to find where you are using the initialized data much more straightforward.
That said, there are a few pitfalls that are outside of the 754 committee's control: as others have noted, not all hardware supports NaN values at speed, which can result in performance hazards. Fortunately, one does not often do a lot of operations on initialized data in a performance-critical setting.
### Solution:2
NaNs are a reasonable choice for a 'no value' sentential (the D programming language uses them for uninitialized values, for instance), but because any comparisons involving them will be false, you can get a few surprises:
• `if (result == DEFAULT_VALUE)`, won't work as expected if `DEFAULT_VALUE` is NaN, as Jon mentioned.
• They can also cause problems with range checking if you're not careful. Consider the function:
` bool isOutsideRange(double x, double minValue, double maxValue) { return x < minValue || x > maxValue; } `
If x is NaN, this function would incorrectly report that x is between minValue and maxValue.
If you just want a magic value for users to test against, I'd recommend positive or negative infinity instead of NaN, as it doesn't come with the same traps. Use NaN when you want it for its property that any operations on a NaN result in a NaN: it's handy when you don't want to rely on callers checking the value, for example.
[Edit: I initially managed to type "any comparisons involving them will be true" above, which is not what I meant, and is wrong, they're all false, apart from NaN != NaN, which is true]
### Solution:3
I have used NaNs in similar situations just because of that: the usual default initialization value 0 is also a valid value. NaNs work fine so far.
It's a good question, by the way, why the default initialization value is usually (for instance, in Java primitive types) 0 and not NaN. Couldn't it as well be 42 or whatever? I wonder what's the rationale of zeros.
### Solution:4
I think it is a bad idea in general. One thing to keep in mind is that most CPU treat Nan much slower then "usual" float. And it is hard to guarantee you will never have Nan in usual settings. My experience in numerical computing is that it often brings more trouble than it worths.
The right solution is to avoid encoding "absence of value" in the float, but to signal it in another way. That's not always practical, though, depending on your codebase.
### Solution:5
Be careful with NaN's... they can spread like wildfire if you are not careful.
They are a perfectly valid value for floats, but any assignments involving them will also equal NaN, so they propagate through your code. This is quite good as a debugging tool if you catch it, however it can also be a real nuisance if you are bringing something to release and there is a fringe case somewhere.
D uses this as rationale for giving floats NaN as default. (Which I'm not sure I agree with.)
### Solution:6
My feelings are that it's a bit hacky, but at least every other numbers you make operations with this NaN value gives NaN as result - when you see a NaN in a bug report, at least you know what kind of mistake you are hunting.
### Solution:7
If your basic need is to have a floating point value which doesn't represent any number which could possibly have been received from the device, and if the device guarantees it will never return NaN, then it seems reasonable to me.
Just remember that depending on your environment, you probably need a special way of detecting NaNs (don't just use `if (x == float.NaN)` or whatever your equivalent is.)
### Solution:8
This sounds like a good use for nans to me. Wish I had thought of it...
Sure, they are supposed to propagate like a virus, that is the point.
I think I would use nan instead of one of the infinities. It might be nice to use a signaling nan and have it cause an event on the first use, but by then its too late it should go quiet on the first use.
### Solution:9
Using NaN as a default value is reasonable.
Note that some expressions, such as (0.0 / 0.0), return NaN.
Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com
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# A question on velocity addition
## Main Question or Discussion Point
Consider frame S1 and another frame S2 moving with a velocity v with respect to S1 along +X direction. Consider a photon comes in -X direction. Velocity of photon in S1 is c.I would like to find the velocity of light in S2 from S1 without forgetting length contraction and time dilation as follows.
Observer in S1 thinks like this. For one second in his clock the clock in S2 will run $$\sqrt{1-v^{2}/c^{2}}$$ seconds. In this time the relative separation between the observer in S2 and photon decreases by (c+v)$$\sqrt{1-v^{2}/c^{2}}$$ considering length contraction.
So from S1 point of view using relativity theory the velocity of photon measured by S2 will be distance/ time = c+v. But it has to be c which we get from velocity addition rule also.
What is wrong here?
Related Special and General Relativity News on Phys.org
jtbell
Mentor
You probably also need to take relativity of simultaneity into account. I haven't thought about your specific situation in detail yet, but in my experience, in general you need to take into account all three of (1) length contraction, (2) time dilation, and (3) relativity of simultaneity, in order to get consistent results in two reference frames.
Or use the Lorentz transformation equations. which incorporate all three effects, and from which all three effects can be derived.
Doc Al
Mentor
Observer in S1 thinks like this. For one second in his clock the clock in S2 will run $$\sqrt{1-v^{2}/c^{2}}$$ seconds. In this time the relative separation between the observer in S2 and photon decreases by (c+v)$$\sqrt{1-v^{2}/c^{2}}$$ considering length contraction.
From the view of the S1 observer, the photon and observer S2 separate at the rate of c+v.
So from S1 point of view using relativity theory the velocity of photon measured by S2 will be distance/ time = c+v. But it has to be c which we get from velocity addition rule also.
No, done correctly, observer S2 will measure the speed of the photon as being c.
What is wrong here?
When taking the results according to S1 and converting them to S2 measurements, you failed to consider the relativity of simultaneity.
[jtbell beat me to it! ]
JesseM
Yes, it's all about the relativity of simultaneity, you have to take into account the idea that each observer measures the difference in time between the beginning and end of the photon's journey using a pair of clocks, one at the position where the photon started and one where it ended, with the two clocks being "synchronized" in that observer's own rest frame. Then you have to realize that two clocks which are a distance x apart and synchronized in their own frame will be out-of-sync by vx/c^2 in a frame where they are moving at speed v along the axis both of them lie on. Here's a numerical example I came up with a while ago showing how if you take into account time dilation, length contraction, and the relativity of simultaneity, you do find that both observers measure the speed of the photon as c:
Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case the relativistic gamma-factor (which determines the amount of length contraction and time dilation) is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.
Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.
Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.
If you want to also consider what happens if, after reaching the front end of the moving ruler at 100 seconds in my frame, the light then bounces back towards the back in the opposite direction towards the back end, then at 125 seconds in my frame the light will be at a position of 75 light-seconds on my ruler, and the back end of the moving ruler will be at that position as well. Since 125 seconds have passed in my frame, 125/1.25 = 100 seconds will have passed on the clock at the back of the moving ruler. Now remember that on the clock at the front read 50 seconds when the light reached it, and the ruler is 50 light-seconds long in its own rest frame, so an observer on the moving ruler will have measured the light to take an additional 50 seconds to travel the 50 light-seconds from front end to back end.
jtbell
Mentor
Aha, I see JesseM got his version in while I was writing mine off-line. Here's mine, anyway:
To make calculations simpler, I measure time in seconds, distance in light-seconds (ls), and speed in ls/sec. In these units, c = 1 ls/sec.
As a "supporting prop", imagine a rod with length L = 10 ls. It is stationary in frame S1, with the left end at x = 0 and the right end at x = 10 ls. It has clocks fastened to its ends, which are synchronized in S1. A light pulse leaves the left end when both clocks read t = 0, and passes the right end when both clocks read t = 10 sec (in S1).
Frame S2 moves to the right at v = 0.8 ls/sec, relative to S1. In this frame, the rod moves to the left at that speed. Because of length contraction, the rod's length is:
$$L^{\prime} = L \sqrt {1 - v^2 / c^2} = 10 \sqrt {1 - 0.8^2} = 6$$
Because of relativity of simultaneity, the two clocks are not synchronized in S2. Their readings differ by the amount
$$\frac {v L} {c^2} = (0.8)(10) = 8$$
sec, with the right-hand clock reading "ahead" of the left-hand clock by this amount. (If the rod were moving to the left, the left-hand clock would be ahead, instead.) Both of these moving clocks "tick" at the same rate (so the difference remains constant), but slower than a stationary clock (in S2).
The light pulse leaves the left end of the rod, going towards the right, when the left-hand clock reads 0 and the right-hand clock reads 8 sec. Let's set the time in S2 so that it's also zero at this moment (t' = 0).
Does the light pulse travel the rod's contracted length L' = 6 ls before it reaches the right end? No, because the right end of the rod is moving to the left! The pulse meets the right end of the rod somewhere between x' = 0 and x' = 6 ls. Let's just call it x', an unknown quantity.
If we know the time t' in S2 when the pulse meets the right end of the rod, we can calculate x' from the initial position of the right end and its speed:
$$x^{\prime} = 6 - 0.8 t^{\prime}$$
But what is t'? It's also unknown, so far.
We do know that when the pulse meets the right end of the rod, the right-hand clock has to read 10 sec. Therefore, during the pulse's flight, 10 - 8 = 2 sec elapse on the (moving) clocks. The time dilation equation gives us the elapsed time on a stationary clock in S2:
$$\Delta t^{\prime} = \frac {\Delta t} {\sqrt{1 - v^2 / c^2}} = \frac {2} {\sqrt{1 - 0.8^2}} = 3.333$$
Therefore the pulse meets the right end of the rod at t' = 3.333 sec, which gives us x' = 6 - (0.8)(3.333) = 3.333.
Now we can calculate the speed of the light pulse in S2 as
$$c^{\prime} = \frac {x^{\prime}} {t^{\prime}} = \frac {3.333} {3.333} = 1 = c$$
This shows that we need all three of length contraction, time dilation, and relativity of simultaneity, in order to make things come out consistently (here, to make the speed of light come out to c = 1 in both frames).
Thanks a lot. I thought relativity of simultaneity was incooperated in length contraction and time dilation.
Or simplify:
The drawing shows the perspective of S2 superposed on S1, with a common event at 0, the left end of the rod. The arc is a geometric means to determine the ratio of t' to t, the prime denoting the moving observer S2. The value t is the S1 time for S2 to travel the length of the rod. It's visually easy to see that the geometry of the S2 frame is 1/g times the S1 frame. For this and similar examples, calculate for the rest frame, then apply the time dilation factor.
v = .8 (c=1), u is the time for light to travel the length of the moving rod.
for S1: x = 10, u = x/(c+v) = 10/1.8 = 5.56 sec
for S2: u' = u/g = 3.33 sec
You are not actually calculating the speed of light, since it's already in the equations you are using. If you calculate distance as c*time, it's a sure thing. SR is formulated to preserve the constant c.
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# Ten Frame
### A 'concrete' model for visualising numbers up to 20.
Learning Objectives:
• YR: Children count reliably with numbers from one to 20, [...] they add and subtract two single-digit numbers and count on or back to find the answer...
• Y1: To represent and use number bonds and related subtraction facts within 20.
• Y1: To solve one-step problems that involve addition and subtraction, using concrete objects and pictorial representations, and missing number problems such as 7 = ? – 9.
A simple model for visualising numbers up to 20. Ideal for '5 and a bit'-style mental visualisations of number.
### Instructions:
You can drag on green, red or white counters. They can be moved to different cells.
The three lines allow you to fill the rest of the cells. This is to prevent teaching delays while you drag on the counters.
To remove a counter simply drag it off the frame.
You can choose from a ten frame, a 20 frame (vertical, for place value) and a 20 frame horizontal
### Examples of use:
Place 8 white counters and 2 red counters in a 10 frame. Ask the children to write as many maths sentences as they can,
8+2=10
2+8=10
10-2=8
10-8=2
5+3+2=10...
Select the vertical 20 frame.
Fill the left-hand frame. Place different numbers of counters in the right-hand frame. Model place value as you create different 'teen numbers'.
Model bonds for 20 e.g. 14 + 6= 20, by using the facts learnt for the bonds for 10. Have fun!
Build on the learning with
these games:
This game is © copyright 2019, J.Barrett, ictgames.com All Rights Reserved.
Made using the and libraries.
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A208502 Number of 4 X n 0..1 arrays avoiding 0 0 0 and 0 0 1 horizontally and 0 1 0 and 1 1 1 vertically. 1
9, 81, 126, 324, 828, 2124, 5436, 13932, 35676, 91404, 234108, 599724, 1536156, 3935052, 10079676, 25819884, 66138588, 169418124, 433972476, 1111644972, 2847534876, 7294114764, 18684254268, 47860713324, 122597730396, 314040583692 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Row 4 of A208501. LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 FORMULA Empirical: a(n) = a(n-1) + 4*a(n-2) for n>4. Conjectures from Colin Barker, Jul 03 2018: (Start) G.f.: 9*x*(1 + 8*x + x^2 - 14*x^3) / (1 - x - 4*x^2). a(n) = (9*2^(-6-n)*((1-sqrt(17))^n*(-109+27*sqrt(17)) + (1+sqrt(17))^n*(109+27*sqrt(17)))) / sqrt(17) for n>2. (End) EXAMPLE Some solutions for n=4: ..0..1..0..0....1..1..1..0....1..0..1..0....1..0..1..1....1..1..0..0 ..0..1..1..0....0..1..0..1....1..1..0..1....1..0..1..1....1..0..1..0 ..1..0..1..0....1..0..1..1....0..1..1..1....0..1..0..0....0..1..1..0 ..1..1..0..1....1..1..1..0....1..0..1..0....1..1..1..1....0..1..0..1 CROSSREFS Cf. A208501. Sequence in context: A209220 A207689 A209225 * A175370 A207695 A158028 Adjacent sequences: A208499 A208500 A208501 * A208503 A208504 A208505 KEYWORD nonn AUTHOR R. H. Hardin, Feb 27 2012 STATUS approved
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Editor selections by Topic and Unit
The Physics Front is a free service provided by the AAPT in partnership with the NSF/NSDL.
## Algebra-Based Physics:Electricity and Electrical Energy Units
Electricity is a natural phenomenon that can be both invisible AND visible, both matter and energy, a type of wave made of protons or a force that cannot be seen. It can move at the speed of light... yet it vibrates in a cord without flowing at all. It can be weightless, or have a small weight. Flowing in a light bulb filament, it transforms into light, but is not used up. It can be stored in batteries. "Electricity" is not only a class of phenomena; it's a type of event.
### A Model for Electricity (5)
#### Activities:
Don't dismiss this resource as too easy for high school. Although simple, this Flash tutorial packs a lot of punch. It goes beyond the commonly-used water flow/electricity analogy to explore what is really happening when electrons flow through a simple circuit. The authors carefully designed it to prevent misconceptions about electric current and prepare students to study electric field and potential.
A simulation that illustrates the "water-flow" model of electricity. The water pump represents a battery in a circuit; coiled water pipe represents a resistor. Students control the rate of flow through the pipe. EDITOR'S NOTE: This model helps students understand some very basic ideas about current flow. We recommend using it with the resource directly above from "The Electricity Book".
Item Type: Interactive Simulation
This high-quality interactive simulation can be easily adapted for both middle school and high school. Students build a virtual DC circuit, using "click and drag" to attach wires, batteries, switches, and resistors. This particular simulation has received excellent reviews in extensive field testing, especially when done in conjunction with a hands-on lab.
Item Type: Interactive Simulation
#### Content Support For Teachers:
It may be easier to understand about electricity if we think of electric charge as a sort of a fluid, like water, as scientists did for over 200 years. NOTE: There are limitations to the fluid analogy. For more, see "The Electricity Book Part 2" under Activities above.
The introductory calculus-based physics curriculum often treats electrostatic and circuit phenomena as separate and unrelated. The problem is that students have little framework for understanding electric field, and often proceed with little sense of the true mechanism of a circuit. This article proposes a new and unified approach for teaching circuits. Students first analyze the behavior of the whole circuit based on dynamic atomic-level models, THEN explore potential and current.
### Electrical Charge (4)
#### Activities:
A set of five experiments that help the students relate current with electric charge. They will build a branching circuit, explore electrolysis in two very different ways, and look at electric charge using capacitors. Each lab has explicit diagrams and tips to help teachers confront misconceptions.
This applet simulates the transient behavior that occurs when a capacitor is being charged and discharged. Students can change the magnitudes of the resistance, capacitance, and the voltage of the battery, as well as flip a switch between its two positions. The applet then graphs both voltage and current in the circuit as functions of time.
#### References and Collections:
This collection of 50+ interactive java tutorials would be an excellent choice to connect physics to "real-world" applications. Designed by well-respected authors, the topics range from simulated magnetic fields and field lines to primers on capacitance, resistance, Ohm's Law, and electromagnetic induction. Included are simulations on how things work, such as vacuum tube diodes, cathode rays, capacitors, AC/DC generators, hard drives, pulsed magnets, and speakers.
#### Content Support For Teachers:
This chapter of an introductory physics textbook provides content support on electrostatics, electric field and potential, current electricity, magnetic field and force, and electromagnetic phenomena. The author, a veteran professor of physics, has summarized his own notes from "lectures that worked" and blended them with calculus-based practice problems with solutions.
### Moving Charges and Electric Circuits (20)
#### Lesson Plans:
If you like the PhET Circuit simulator (and who doesn't?) you'll appreciate this series of three inquiry-based activities developed especially to accompany the sim. Designed by PhET Gold-Star winner Trish Loeblein, the activities explore basic properties of electric circuits, resistors in series and parallel circuits, and more. Bonus: includes clicker questions.
Don't have time to do three lessons with the PhET Circuit Simulator? This lesson plan, specifically designed for use with the simulation, takes one day in computer lab and one follow-up day. We highly recommend it as a way to let your students explore circuits, learn from mistakes, and be better prepared to participate in the hands-on circuit lab.
This one-day lab is a great way for students to investigate factors causing short circuits. Reproducible prediction charts help students learn by gauging their preconceived ideas against observed outcomes in the lab. Materials are readily accessible and inexpensive to obtain.
This stand-out lesson from the Institute of Electrical & Electronics Engineers gives step-by-step instructions for designing and building a voltage divider -- a form of linear circuit capable of producing a wide range of output voltages. Students explore the mathematical relationships of parallel & series resistors as they build LED's. It is intended as the 3rd in a set of circuit lessons by the same publisher.
Item Type: Inquiry-Based Lesson
Level: High School Physics
Duration: Two Class Periods
#### Activities:
This high-quality interactive simulation is a good choice for beginning high school physics or physical science. Students build a virtual DC circuit, using the mouse to attach wires, batteries, switches, and resistors. This particular simulation has received excellent reviews in extensive field testing, especially when done in conjunction with a hands-on lab. See Lesson Plans above for a recommended lesson developed specifically for use with this simulation.
This applet simulates the behavior of a simple RLC circuit with an AC voltage source.
This item offers guided practice to students in designing a variety of circuits, with pop-up explanations to augment simulated images.
This applet simulates the transient behavior of a simple RC circuit. Students can change the magnitudes of the resistance, capacitance, and the voltage of the battery, as well as flip a switch between its two positions. The applet then graphs both voltage and current in the circuit as functions of time.
A collection of 8 experiments designed to introduce important concepts of electricity to beginning students. Each lab is accompanied by instructional tips to help students form a solid basis for a future study of resistance, Ohm's Law, and potential difference.
#### References and Collections:
This page offers detailed background information on 15 different types of batteries. The author discusses the historical background of each battery, augmented with cross-sectional images and full explanations of how the batteries function.
Want to go beyond your traditional textbook in a unit on electric circuits? This free web-based textbook offers solid content support for both learners and teachers. The text is student-friendly, blended with many diagrams and photos. Each section is further supplemented with suggested labs and activities. Did we mention, it's free?
#### Content Support For Teachers:
It may be easier to understand about electricity if we think of electric charge as a sort of a fluid, like water, as scientists did for over 200 years.
This item is an educator's guide featuring easily-understood content support in fundamentals of electricity, electrostatic charge, circuit basics, conductors, and insulators. Labs and assessments are also included.
This interactive applet displays the phasor diagram for an AC circuit. Probably best used within a teacher-directed activity, it is intended to show current through a circuit as a function of frequency. By controlling the frequency, resistance, capacitance, and inductance of the circuit, students can conceptualize the relationship of these variables and better understand how to interpret the related graphs.
This item is a unit of study on electric circuits for beginning high school physics. Especially noteworthy are the detailed labs that encourage the process of inquiry and promote critical thinking. Other included resources are numerical problems, puzzles, reference materials, and practice test questions/answers.
This is a research article investigating high school and college students' understanding of how DC circuits work. The analysis indicates that students, especially females, tend to hold multiple misconceptions, even after instruction. The main source of misconception, as reported by the article, is with confusion about the underlying mechanism of electric circuits and the meaning of "current".
#### Student Tutorials:
This item is a concise explanation for beginning physics students on how batteries work. It includes a brief history of the battery, the roles of Galvani and Volta in developing an early battery, and diagrams to explain how current flows in a battery.
Good jumping-off point for students with little background in electricity. This is Chapter 1 of free online textbook, All About Circuits. With entertaining language and detailed diagrams, the author helps students form accurate concepts of electron transfer and charge interaction so they can successfully apply the knowledge in a lab.
This animated tutorial does a great job to promote understanding of current: what it is, how it is produced, and how it moves. The animations show students the difference between DC and AC current flow, and explain how and AC generator works. Especially recommended for classes where Internet technology is limited: the tutorial is delivered in HTML format.
#### Assessment:
This is an exemplary set of 50+ short assessments on the basics of electricity and electric circuits. They offer teachers flexibility: 1) students can go online to interactively answer questions and see correct responses, 2) Students can work in small groups to discuss strategies before looking at answers, or 3) Teachers can print the worksheets with answers hidden. Most of the questions require analytical reasoning and will help teachers gauge whether students are getting the big picture.
### Electric Force and Coulomb's Law (2)
#### Activities:
Coulomb's Law is used to calculate the electric force between charges. To help students grasp the underlying concepts, this simulation lets them drag a moveable charge to see the connection between the force vector and proximity to the fixed charge.
#### Content Support For Teachers:
It may be easier to understand about electricity if we think of electric charge as a sort of a fluid, like water, as scientists did for over 200 years.
### Resistance and Ohm's Law (6)
#### Lesson Plans:
This stand-out lesson from the Institute of Electrical & Electronics Engineers gives step-by-step instructions for designing and building a voltage divider -- a form of linear circuit capable of producing a wide range of output voltages. Students explore the mathematical relationships of parallel & series resistors as they build LED's. It is intended as the 3rd in a set of circuit lessons by the same publisher.
Item Type: Inquiry-Based Lesson
Level: High School Physics
Duration: Two Class Periods
#### Activities:
This java simulation demonstrates the relationship among current, voltage, and resistance. Students adjust resistance and voltage up or down in a simple circuit and watch the results on a simulated ammeter.
As students adjust current flow in this Java simulation, the resulting changes are represented on both a macro and nanoscale. This activity helps the beginner understand how electron collision causes resistance. Try teaming this simulation with the one above on Ohm's Law.
Robust activity features six molecular models to explore relationships among voltage, current, and resistance. This well-sequenced resource will help learners understand how current is different from voltage and visualize how electron movement is related to conductivity. More advanced students can explore a hydrogen fuel cell model, an incandescent light bulb filament, and electromotive force.
Item Type: Interactive Model
Duration: Two Class Periods
In this activity from the PTRA manual "Role of the Laboratory in Teaching Introductory Physics", students set up their first circuit using meters and specially made resistors in heat sink boxes, which do not require alligator clips and don't burn hands.
Item Type: Laboratory
Duration: 1-2 Class Periods
#### Content Support For Teachers:
This page is a comprehensive tutorial on resistance, accompanied by three interactive simulations. One allows students to glimpse resistance at a molecular level; the second explores Ohm's Law; and the third introduces the use of color coding in composition resistors.
### Sources of Electrical Energy (2)
#### Content Support For Teachers:
This page offers detailed background information on 15 different types of batteries. The author discusses the historical background of each battery, augmented with cross-sectional images and full explanations of how the batteries function.
#### Student Tutorials:
This item is a concise explanation for beginning physics students on how batteries work. It includes a brief history of the battery, the roles of Galvani and Volta in developing an early battery, and diagrams to explain how current flows in a battery.
### Applying Concepts of Electricity (7)
#### Lesson Plans:
A very effective lab for reinforcing the importance of circuit continuity. A lamp is connected to a battery with jumper wires. After measuring normal voltages in a functioning circuit, students break the circuit at each of the four connecting points, then measure again. Editor's Note: For additional practice in measuring voltage, see "Voltmeter Usage" activity below.
A collection of 8 experiments designed to introduce important concepts of electricity to beginning students. Each lab is accompanied by instructional tips to help students form a solid basis for a future study of resistance, Ohm's Law, and potential difference.
#### Activities:
A very good introduction to the multimeter, an electronic instrument that measures voltage, current, and resistance. It will help students become comfortable using either a digital or analog multimeter with batteries, an LED, and a simple "hobby" motor.
This is a companion lab to the resource directly above, "Voltmeter Usage". It helps students get practice in using the ammeter function of a multimeter to measure current -- the rate of electron flow in a circuit. Detailed instructions and photos make set-up easy.
A set of 5 creative labs for constructing various types of primary battery cells, all appropriate for use in the high school science classroom. Two of the labs can be performed without corrosive chemicals: the voltaic pile and the lemon cell. The remaining 3 labs use either sulfuric acid or zinc chloride. Each lab contains detailed information on safety precautions and classroom set-up.
Students learn about systems as they reverse-engineer a disposable camera containing both electrical and mechanical components. They create a systems diagram for the deconstructed device. Then, in partnership with other participants across the country, learners will reassemble the device and test their reconstruction against quality controls. The project is free with teacher registration.
Item Type: Digital Telecollaboration
Duration: Multi-Day
#### Assessment:
This is an exemplary set of 50+ short assessments on the basics of electricity and electric circuits. They offer teachers flexibility: 1) students can go online to interactively answer questions and see correct responses, 2) Students can work in small groups to discuss strategies before looking at answers, or 3) Teachers can print the worksheets with answers hidden. Most of the questions require analytical reasoning and will help teachers gauge whether students are getting the big picture.
### Electricity: A Historical Perspective (8)
#### Activities:
In this Java simulation, your students play with a replication of Coulomb's historic torsion balance: a device used to measure electric force between charges. Coulomb's methodical measuring laid the foundation for Coulomb's Law, a fundamental principle of electricity and magnetism.
It sometimes helps students with concept formation if they can see how early scientists made momentous discoveries. In this tutorial, students play with a simulation of the voltaic pile device invented in 1800 by Volta -- commonly known as the world's first battery. Battery cells can still be assembled using this "recipe". SEE ITEM DIRECTLY BELOW for a lab to construct the voltaic pile device in the classroom.
If you'd like your students to replicate Volta's groundbreaking experiment with the voltaic pile device (the world's first battery), here is a lesson plan (scroll down to Page 2 of the document). They will construct their own voltaic pile batteries and get a better understanding of how electrochemical reactions work. No harsh chemicals or safety hazards.
#### References and Collections:
In this excellent set of 50+ short biographies, kids can read about the challenges of early inventors AND follow links to simulations of the devices they invented. For example, the Coulomb biography offers a simulation of the torsion balance; the Volta biography has a simulated voltaic pile battery.
A short biography of Charles-Augustin de Coulomb, whose historic work in the 18th century was essential to the development of electromagnetic theory. Read about his experiments with a torsion balance and how he discovered the mathematical relationship known as Coulomb's Law.
#### Student Tutorials:
In the 18th-century, Italian scientist Alessandro Volta proposed the theory that electrical current is generated by contact between different metals. His experimental work resulted in the "voltaic pile" battery, the first known source of sustainable electric current. For a simulation of the voltaic pile device, see "Activities" above.
This is a short biography of Charles Augustin de Coulomb, the 18th-century scientist whose experiments with a torsion balance gave rise to Coulomb's Law -- a fundamental principle of physics that defines the electrical force between two charged particles as a predictable mathematical relationship. For a simulation of Coulomb's torsion balance, see "Activities" above.
German physicist Georg Ohm was different, and his fellows at the time were not too supportive. This early giant in the field of electricity took a mathematical approach to electric current, at a time when his peers relied almost exclusively on lab experimentation. His perseverance resulted in Ohm's Law, which clarified the relationship between electrical current, resistance, and voltage. This is his biography.
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# Bresenham line drawing algorithm example ppt
### Computer Graphics 4Bresenham Line Drawing Algorithm
Bresenham’S Circle Drawing AlgorithmauthorSTREAM. Line Drawing 15 Scan-Conversion Algorithms Drawing a Line Bresenhamʼs Algorithm: Example 38, Line Drawing 15 Scan-Conversion Algorithms Drawing a Line Bresenhamʼs Algorithm: Example 38.
### Tech-Algorithm.com ~ Drawing Line Using Bresenham Algorithm
Bresenham’s Line and Circle Algorithms. Bresenham's circle algorithm is derived from the midpoint circle algorithm. It usually comes after drawing the Just as with Bresenham's line algorithm,, ... Bresenham Line Drawing Algorithm, Circle Drawing - PowerPoint PPT Presentation Bresenham Example (cont) Bresenham Line Drawing Algorithm,.
Numerical on Bresenham's line drawing algorithm with step by step form.. Bresenham's Line Algorithm by studying the visualization. example of a button. Bresenham's Lines Algorithm Visualization Using Flash
Computer Graphics 4:Bresenham Line Drawing Algorithm ,Ask Latest information,Abstract,Report,Presentation (pdf,doc,ppt),Computer Graphics 4:Bresenham Line Drawing Bresenham’s Line Algorithm. The Problem (cont…). What happens when we try to draw this on a pixel based display?. How do we choose which pixels to turn on?.
Bresenham’s Circle drawing algorithm 2 – r 2 If p k < 0 , this midpoint is inside the circle and the pixel on the scan line y k PowerPoint Presentation: Bresenham Line Drawing Algorithm, Circle Drawing & Polygon Filling. 2 of 39 Contents Mid-Point Circle Algorithm Example To see the mid-point circle algorithm in
Bitmap/Bresenham's line algorithm The following program tests the above bresenham function by drawing 100 lines into an image and visualizing Usage example: Bresenham’s Line Algorithm. The Problem (cont…). What happens when we try to draw this on a pixel based display?. How do we choose which pixels to turn on?.
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Bresenham Circle Drawing Algorithm,. Contents. In today’s lecture we’ll have a look at: Bresenham’s Circle drawing algorithm Exercise using Bresenham’s Slide 1 Drawing Lines The Bresenham Algorithm for drawing lines and filling polygons Slide 2 Plotting a line-segment Bresenham published algorithm in 1965 It was
Draw a Line Using Bresenham Draw a Line Using Bresenham Line Algorithm - Notes \\BGI"); outtextxy(150,10, "Bresenham line drawing algorithm 1. 2 of 60 Towards the Ideal Line We can only do a discrete approximation Illuminate pixels as close to the true path as possible, consider bi-level display only
Scan-Line Algorithm Aliasing / Antialiasing Examples ¥ Line drawing algorithms such as Bresenham's can easily be modified to Recall: Bresenham’s Line‐Drawing Algorithm Bresenham’s Line‐Drawing Algorithm Example: to find line segment between lecture25.ppt [Compatibility Mode]
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A Fast Bresenham Type Algorithm For Drawing Ellipses by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 C Program for BRESENHAM’S ELLIPSE DRAWING ALGORITHM. Mohanraj 1 comment . To draw an ellipse using Bresenham ellipse drawing algorithm. ALGORITHM: 1. Start. 2.
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Bresenhams Line Drawing Algorithm Bresenhams Line Drawing Algorithm Bresenhams from COMPUTER S CSN-372 at Indian Institute of Technology, Roorkee Line Drawing Algorithms - Bresenham -. Assist. Prof. Dr. Ahmet Sayar Computer Engineering Department Computer Graphics Course Kocaeli University Fall 2013.
### Recall Bresenham’s Line Drawing Algorithm WPI
Bresenham’S Circle Drawing AlgorithmauthorSTREAM. Line Drawing 15 Scan-Conversion Algorithms Drawing a Line Bresenhamʼs Algorithm: Example 38, Bresenham's line algorithm is an As an example, the line Patrick-Gilles Maillot's Thesis an extension of the Bresenham line drawing algorithm to.
Bresenham’s Line Generation Algorithm GeeksforGeeks. Bresenham's circle/ellipse drawing algorithm. Here the general approach is based on lines drawing and sqrt function. Using Bresenham's approach we exclude, Bresenham’s Line Algorithm Bresenham’s algorithm is a highly efficient incremental method Bresenham's line algo. Example 1 Scan convert a line from (1,1.
### DDA line drawing algorithm explanation with example
Bresenham's Algorithm University of California Davis. Computer Graphics Scan Bresenham’s Algorithm (Jack Bresenham 1965) By drawing from both ends of line at once 4 pixels can be chosen with one discriminator. https://en.m.wikipedia.org/wiki/Line_drawing_algorithm Bitmap/Bresenham's line algorithm The following program tests the above bresenham function by drawing 100 lines into an image and visualizing Usage example:.
The Bresenham Algorithm for drawing lines on the discrete plane, such as computer monitor is one of the fundamental algorithms in computer graphics. General Bresenham Line Drawing Algorithm. Buscar Buscar
Bresenham's line drawing algorithm & Mid Point Circle algorithm graphics on the Bresenham line drawing algorithm, the midpoint of the algorithm to achieve Circle, and pentagons, hexagons, heptagon filled....
Bresenham’s line algorithm is an algorithm that determines the points of an n-dimensional raster that should be selected in order to form a close approximation to a Bresenham's circle algorithm is derived from the midpoint circle algorithm. It usually comes after drawing the Just as with Bresenham's line algorithm,
advantages and disadvantages of bresenham s About advantages and disadvantages of bresenham s algorithm is Not bresenham line drawing algorithm ppt, DERIVATION OF THE BRESENHAM’S LINE ALGORITHM Assumptions : input: line endpoints at (X1,Y1) and (X2, Y2) Bresenham_derivation Author: exc Created Date:
Examples on line drawing algorithms: 1. DDA algorithm 2. Bresenham's algorithm Bresenham’s Line Algorithm Bresenham’s algorithm is a highly efficient incremental method Bresenham's line algo. Example 1 Scan convert a line from (1,1
Drawing a circle on the screen is a little complex than drawing a line. There are two popular algorithms for generating a circle − Bresenham’s Algorithm and Bresenham’s Midpoint Algorithm How to Draw a Line? 1. Compute slope 2. Microsoft PowerPoint - Wk 2 Lec02_Bresenham Author: hansen
Computer Graphics 5: Line Drawing Algorithms . 2 of 32 determined for the line in our previous example: (2, 2), (3, 2. 3 / 5 Bresenham’s Line Drawing Algorithm Bresenham’s Line and Circle Algorithms Bresenham’s Midpoint Circle Algorithm. For drawing circles, we could easily develop an algorithm that makes
Computer Graphics 5: Line Drawing Algorithms . 2 of 32 determined for the line in our previous example: (2, 2), (3, 2. 3 / 5 Bresenham’s Line Drawing Algorithm Bresenham’s Line Algorithm Bresenham’s algorithm is a highly efficient incremental method Bresenham's line algo. Example 1 Scan convert a line from (1,1
graphics on the Bresenham line drawing algorithm, the midpoint of the algorithm to achieve Circle, and pentagons, hexagons, heptagon filled.... BRESHENHAM’S ALGORITHM Bresenham’s Algorithm Consider a line with initial point (x 1,y 1) example, in which we wish to draw a line from (0,0)
Illustration of the result of Bresenham's line algorithm. As an example, the line then this extension of the Bresenham line drawing algorithm to perform BRESHENHAM’S ALGORITHM Bresenham’s Algorithm Consider a line with initial point (x 1,y 1) example, in which we wish to draw a line from (0,0)
## CS 543 Computer Graphics Lecture 9 (Part I) Raster
Line Characterizations Bresenham’s Midpoint Algorithm. We are given vertices and we need to use Bresenham's Line algorithm to draw the lines run your code on 2 simple examples, Bresenham's Line Drawing Algorithm. 13., ... high quality dda line drawing algorithm dda line algorithm in computer graphics with example ppt; dda line algorithm in bresenham line algorithm example;.
### MATH 308 Project The Bresenham Line-Drawing Algorithm
Computer Graphics Line Generation Algorithm Tutorials Point. Bresenham’s Line Generation Algorithm. Below are some assumptions to keep algorithm simple. We draw line from //en.wikipedia.org/wiki/Bresenham’s_line, Bresenham’s Line Generation Algorithm. Below are some assumptions to keep algorithm simple. We draw line from //en.wikipedia.org/wiki/Bresenham’s_line.
Write Short Note on Digital Differential Analyzer axis by using the slope equation of the line. For example if the end Bresenham’s Line Drawing Algorithm. Bresenham’s Line Algorithm Bresenham’s algorithm is a highly efficient incremental method Bresenham's line algo. Example 1 Scan convert a line from (1,1
The basic "line drawing" algorithm used in computer graphics is Bresenham's Algorithm. table to see the complete operation of the algorithm on this example. This algorithm was developed by Jack E. Bresenham in 1962 at IBM. As stated above, in this lecture, I will explain how to draw lines using the Bresenham's line
The Bresenham algorithm is another incremental scan conversion algorithm. It is useful alternative for the DDA The big advantage of this algorithm is that it u… Bresenham Circle Drawing Algorithm,. Contents. In today’s lecture we’ll have a look at: Bresenham’s Circle drawing algorithm Exercise using Bresenham’s
Bresenham Line Drawing Algorithm, Circle Drawing & Polygon Filling. 2 of 39 Contents Mid-Point Circle Algorithm Example To see the mid-point circle algorithm in Line Drawing 15 Scan-Conversion Algorithms Drawing a Line Bresenhamʼs Algorithm: Example 38
Line and Curve Drawing Algorithms. Line Drawing. y = m . x b. m Bresenhams Line Algorithm yk1 yk xk xk1 yk1 du y dl yk xk Example. 23 Antialiasing . Line Drawing - Introduction to Computer Graphics • No change to line drawing algorithm 4. Introduction to Computer Graphics - Lecture Slides.
Recall: Bresenham’s Line‐Drawing Algorithm Bresenham’s Line‐Drawing Algorithm Example: to find line segment between lecture25.ppt [Compatibility Mode] DERIVATION OF THE BRESENHAM’S LINE ALGORITHM Assumptions : input: line endpoints at (X1,Y1) and (X2, Y2) Bresenham_derivation Author: exc Created Date:
Line Drawing - Introduction to Computer Graphics • No change to line drawing algorithm 4. Introduction to Computer Graphics - Lecture Slides. advantages and disadvantages of bresenham s About advantages and disadvantages of bresenham s algorithm is Not bresenham line drawing algorithm ppt,
Bresenham’s Line and Circle Algorithms Bresenham’s Midpoint Circle Algorithm. For drawing circles, we could easily develop an algorithm that makes Bresenham’s Circle drawing algorithm 2 – r 2 If p k < 0 , this midpoint is inside the circle and the pixel on the scan line y k PowerPoint Presentation:
The basic Bresenham algorithm Consider drawing a line on a raster grid where we restrict the allowable slopes of the line to the range . If we further restrict the 1. 2 of 60 Towards the Ideal Line We can only do a discrete approximation Illuminate pixels as close to the true path as possible, consider bi-level display only
Draw a Line Using Bresenham Draw a Line Using Bresenham Line Algorithm - Notes \\BGI"); outtextxy(150,10, "Bresenham line drawing algorithm General Bresenham Line Drawing Algorithm. Buscar Buscar
Draw a Line Using Bresenham Draw a Line Using Bresenham Line Algorithm - Notes \\BGI"); outtextxy(150,10, "Bresenham line drawing algorithm ... 2∆x Repeat step 4 (Δx ) times Bresenham ’s Line Drawing Algorithm Bresenham’s Line Drawing Example To illustrate PowerPoint Presentation Author:
Line Drawing 15 Scan-Conversion Algorithms Drawing a Line Bresenhamʼs Algorithm: Example 38 Bresenham’s Circle drawing algorithm 2 – r 2 If p k < 0 , this midpoint is inside the circle and the pixel on the scan line y k PowerPoint Presentation:
BRESHENHAM’S ALGORITHM Bresenham’s Algorithm Consider a line with initial point (x 1,y 1) example, in which we wish to draw a line from (0,0) A Fast Bresenham Type Algorithm For Drawing Ellipses by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405
Line and Curve Drawing Algorithms. Line Drawing. y = m . x b. m Bresenhams Line Algorithm yk1 yk xk xk1 yk1 du y dl yk xk Example. 23 Antialiasing . Write Short Note on Digital Differential Analyzer axis by using the slope equation of the line. For example if the end Bresenham’s Line Drawing Algorithm.
Bresenham Line Drawing Algorithm Circle Drawing & Polygon Filling,Ask Latest information,Abstract,Report,Presentation (pdf,doc,ppt), Bresenham Line Drawing Algorithm Bresenham's circle/ellipse drawing algorithm. Here the general approach is based on lines drawing and sqrt function. Using Bresenham's approach we exclude
Bresenham's circle algorithm is derived from the midpoint circle algorithm. It usually comes after drawing the Just as with Bresenham's line algorithm, Here you will get program for bresenham's line drawing algorithm in C Bresenham’s Midpoint Circle Algorithm in C i.e for example if we give input where
In Computer Graphics the first basic line drawing algorithm is you can draw a line. Also Read: Bresenham’s Line URL Example Best Fit Algorithm in Bresenham Line Drawing Algorithm, Circle Drawing & Polygon Filling. 2 of 39 Contents Mid-Point Circle Algorithm Example To see the mid-point circle algorithm in
Bresenham’s Line and Circle Algorithms Bresenham developed a circle drawing algorithm that does exactly this, using mostly integer arithmetic, as follows. ... y yk yk+1 xk+1 dlower dupper The Bresenham Line Algorithm BRESENHAM’S LINE DRAWING ALGORITHM Input the two Bresenham Example Bresenham Example
... bresenham line drawing algorithm ppt bresenham line algorithm in computer graphics explain bresenham's line drawing algorithm with example bresenham line drawing We are given vertices and we need to use Bresenham's Line algorithm to draw the lines run your code on 2 simple examples, Bresenham's Line Drawing Algorithm. 13.
The Bresenham Line Algorithm algorithm for drawing such a line is the Bresenham Line Here’s an example of what this will look like: A line in a raster 4. any line drawing algorithm. In this work the value of depth is determined for each that are required, for example, in Bresenham’s algorithm. Experimental
Bresenham’s Circle drawing algorithm 2 – r 2 If p k < 0 , this midpoint is inside the circle and the pixel on the scan line y k PowerPoint Presentation: Line and Curve Drawing Algorithms. Line Drawing. y = m . x b. m Bresenhams Line Algorithm yk1 yk xk xk1 yk1 du y dl yk xk Example. 23 Antialiasing .
### Bresenham’s Line Algorithm Technical PDF
Bresenham's line algorithm IPFS. ... bresenham line drawing algorithm ppt bresenham line algorithm in computer graphics explain bresenham's line drawing algorithm with example bresenham line drawing, We are given vertices and we need to use Bresenham's Line algorithm to draw the lines run your code on 2 simple examples, Bresenham's Line Drawing Algorithm. 13..
Parallel algorithms for line generation Springer. Write Short Note on Digital Differential Analyzer axis by using the slope equation of the line. For example if the end Bresenham’s Line Drawing Algorithm., Line and Curve Drawing Algorithms. Line Drawing. y = m . x b. m Bresenhams Line Algorithm yk1 yk xk xk1 yk1 du y dl yk xk Example. 23 Antialiasing ..
### Bresenhams line drawing algorithm bresenhams line
The Bresenham Line Algorithm Tom Carter. In Computer Graphics the first basic line drawing algorithm is you can draw a line. Also Read: Bresenham’s Line URL Example Best Fit Algorithm in https://en.wikipedia.org/wiki/Midpoint_circle_algorithm Line and Curve Drawing Algorithms. Line Drawing. y = m . x b. m Bresenhams Line Algorithm yk1 yk xk xk1 yk1 du y dl yk xk Example. 23 Antialiasing ..
The basic "line drawing" algorithm used in computer graphics is Bresenham's Algorithm. table to see the complete operation of the algorithm on this example. CS 543: Computer Graphics Lecture 9 (Part I): Bresenham’s Line-Drawing Algorithm n Example: Bresenham’s Line-Drawing Algorithm
The Bresenham Algorithm for drawing lines on the discrete plane, such as computer monitor is one of the fundamental algorithms in computer graphics. Bresenham's line drawing algorithm & Mid Point Circle algorithm
Bresenham Line Drawing Algorithm, Circle Drawing & Polygon Filling. 2 of 39 Contents Mid-Point Circle Algorithm Example To see the mid-point circle algorithm in Scan-Line Algorithm Aliasing / Antialiasing Examples ¥ Line drawing algorithms such as Bresenham's can easily be modified to
27/01/2014 · Prob with Midpoint Circle Algorithm Example [PPT] View web2.aabu.edu.jo Output Primitives; Line Drawing Algorithms. Bresenham’s Midpoint Algorithm How to Draw a Line? 1. Compute slope 2. Microsoft PowerPoint - Wk 2 Lec02_Bresenham Author: hansen
... bresenham line drawing algorithm ppt bresenham line algorithm in computer graphics explain bresenham's line drawing algorithm with example bresenham line drawing advantages and disadvantages of bresenham s About advantages and disadvantages of bresenham s algorithm is Not bresenham line drawing algorithm ppt,
CS 543: Computer Graphics Lecture 9 (Part I): Bresenham’s Line-Drawing Algorithm n Example: Bresenham’s Line-Drawing Algorithm Bresenham's circle algorithm is derived from the midpoint circle algorithm. It usually comes after drawing the Just as with Bresenham's line algorithm,
Illustration of the result of Bresenham's line algorithm. As an example, the line then this extension of the Bresenham line drawing algorithm to perform 26/06/2017 · DDA line drawing algorithm explanation with example with PowerPoint Derivation & Explanation of Bresenham's Line Drawing Algorithm in
Bresenham’s Line and Circle Algorithms Bresenham developed a circle drawing algorithm that does exactly this, using mostly integer arithmetic, as follows. Bresenham Line Drawing Algorithm Circle Drawing & Polygon Filling,Ask Latest information,Abstract,Report,Presentation (pdf,doc,ppt), Bresenham Line Drawing Algorithm
A Fast Bresenham Type Algorithm For Drawing Ellipses by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 Bresenham's line drawing algorithm & Mid Point Circle algorithm
Bresenham’s Line and Circle Algorithms Bresenham developed a circle drawing algorithm that does exactly this, using mostly integer arithmetic, as follows. Bresenham Circle Drawing Algorithm,. Contents. In today’s lecture we’ll have a look at: Bresenham’s Circle drawing algorithm Exercise using Bresenham’s
Numerical on Bresenham's line drawing algorithm with step by step form.. DERIVATION OF THE BRESENHAM’S LINE ALGORITHM Assumptions : input: line endpoints at (X1,Y1) and (X2, Y2) Bresenham_derivation Author: exc Created Date:
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Home > Confidence Interval > Standard Error Proportion Difference
# Standard Error Proportion Difference
## Contents
If the sample sizes are equal (n1 = n2 = n), then . A pilot sample which is drawn from the population and used as an estimate of . 2. in mathematics from the University of Notre Dame. The rules for inference about two proportions firmly go both(!) ways. http://comunidadwindows.org/confidence-interval/standard-error-of-proportion-difference.php
Determination of the sample size for estimating proportions The manner of finding sample sizes for estimating a population proportion is basically the same as for estimating a mean. And since each population is more than 20 times larger than its sample, we can use the following formula to compute the standard error (SE) of the difference between proportions: SE The Variability of the Difference Between Proportions To construct a confidence interval for the difference between two sample proportions, we need to know about the sampling distribution of the difference. Suppose your random sample of 100 females includes 53 females who have seen an Elvis impersonator, so is 53 divided by 100 = 0.53. http://stattrek.com/estimation/difference-in-proportions.aspx?Tutorial=AP
## Confidence Interval For Difference In Proportions Calculator
When we carry out a test with null hypothesis p1 = p2, all our calculations are based on the assumption that this null is true -- so our best estimate for Using a simple random sample, they select 400 boys and 300 girls to participate in the study. standard-error proportion share|improve this question edited Jan 14 at 15:54 asked Jan 14 at 15:18 C8H10N4O2 318116 Am I overthinking this? The range of the confidence interval is defined by the sample statistic + margin of error.
Suppose we classify choosing Superman as a success, and any other response as a failure. To find a confidence interval for the average difference between these two populations we compute$\text{Standard Error for Difference} = \sqrt{0.103^{2}+0.465^{2}} \approx 0.476$If we think about all possible ways to draw a Find the sample proportion for the first sample by taking the total number from the first sample that are in the category of interest and dividing by the sample size, n1. 2 Proportion Z Interval Example Both samples should be independent.
The standard error is estimated by the formula: Confidence interval The 100(1- ) percent confidence interval for - is given by: Interpretation of the interval The Standard Error Two Proportions Calculator We can't estimate from a value of ; we need to go back to the data and look at deviations. However, the 8% difference is based on random sampling, and is only an estimate of the true difference. http://davidmlane.com/hyperstat/B73789.html So our estimate of p1 - p2 is .
And the uncertainty is denoted by the confidence level. Confidence Interval For Two Population Proportions Calculator From the Normal Distribution Calculator, we find that the critical value is 1.645. Use the sample proportions (p1 - p2) to estimate the difference between population proportions (P1 - P2). Results of the Smoking and wrinkles study (example 10.6) SmokersNonsmokersSample Size150250Sample Proportion with Prominent Wrinkles95/150 = 0.63105/250 = 0.42Standard Error for Proportion$$\sqrt{\frac{0.63(0.37)}{150}} = 0.0394$$$$\sqrt{\frac{0.42(0.58)}{250}} = 0.0312$$How do the smokers compare to
## Standard Error Two Proportions Calculator
When each sample is small (less than 5% of its population), the standard deviation can be approximated by: SEp1 - p2 = sqrt{ [p1 * (1 - p1) / n1] + A 95% confidence interval for the difference in proportions p1-p2 is or . Confidence Interval For Difference In Proportions Calculator Substituting this value of for both p1 and p2 gives our estimate of ; we have merged the data from the two samples to obtain what is called the "pooled" estimate The Confidence Interval For The Difference Between Two Independent Proportions Lesson 11: Hypothesis Testing Lesson 12: Significance Testing Caveats & Ethics of Experiments Reviewing for Lessons 10 to 12 Resources References Help and Support Links!
We say that we are 95% confident that the difference between the two population proportions, - , lies tbetweenhe calculated limits since, in repeated sampling, about 95% of the intervals constructed this content Then, we have plenty of successes and failures in both samples. Use the sample proportions (p1 - p2) to estimate the difference between population proportions (P1 - P2). Select a confidence level. 2 Proportion Z Interval Conditions
If an upper limit is suspected or presumed, it could be used to represent p. 2. Identify a sample statistic. Find and divide that by n2. http://comunidadwindows.org/confidence-interval/standard-error-2-proportion.php Identify a sample statistic.
Orton, Scott AdamsList Price: $9.99Buy Used:$0.01Buy New: $1.77Texas Instruments TI-86 Graphing CalculatorList Price:$150.00Buy Used: $23.00Approved for AP Statistics and Calculus About Us Contact Us Privacy Terms of Use Confidence Interval Difference In Proportions Ti-84 We cannot compare the left-hand results and the right-hand results as if they were separate independent samples. From the Normal Distribution Calculator, we find that the critical value is 1.645. ## Similarly, find for the second sample. Thus, the sample statistic is pboy - pgirl = 0.40 - 0.30 = 0.10. Data from a study of 60 right-handed boys under 10 years old and 60 right-handed men aged 30-39 are shown in Table 10.3.Table 10.3 Grip Strength (kilograms) Average and Standard Deviation D) Confidence interval for the difference of two population proportions When studying the difference between two population proportions, the difference between the two sample proportions, - , can be used as Margin Of Error For Two Proportions Calculator The standard deviation of any variable involves the expression . Why do I even need a confidence interval?" All those two numbers tell you is something about those 210 people sampled. We pool for the one case, and do not pool for the others, because in the one case we must treat the two sample proportions as estimates of the same value Texas Instruments TI-83-Plus Silver EditionList Price:$169.99Buy Used: $49.98Buy New:$55.00Approved for AP Statistics and CalculusSchaums Outline of Statistics, Fourth Edition (Schaum's Outline Series)Murray Spiegel, Larry StephensList Price: $19.00Buy Used:$0.01Buy check over here Select a confidence level.
Suppose we classify choosing Superman as a success, and any other response as a failure. Find standard deviation or standard error. You estimate the difference between two population proportions, p1 - p2, by taking a sample from each population and using the difference of the two sample proportions, plus or minus a That is, we are 90% confident that the true difference between population proportion is in the range defined by 0.10 + 0.06.
Getting around copy semantics in C++ What to do when majority of the students do not bother to do peer grading assignment? The approach that we used to solve this problem is valid when the following conditions are met. Star Fasteners Is it possible to fit any distribution to something like this in R?
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phys1cquiz3
# phys1cquiz3 - Chapter 27 Wiens Displacement Law max(T =...
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Chapter 27 Wien’s Displacement Law: λ max (T) = 0.2898 x 10^-2 (m x K) where λ max is the wavelength at which the curve peaks and T is the absolute temperature of the object emitting the radiation Classical theory says that predicts that intensity of blackbody radiation should be infinite but experimental data shows that is should approach zero ultraviolet catastrophe Resonators have discrete energies E n , given by E n = nhf where n is a positive integer called a quantum number, f is the frequency of vibration of the resonator, and h is Planck’s constant, h = 6.626 x 10 -34 (J x s) When V is equal to or more negative than - V s , the stopping potential, no electrons reach C and the current is zero. The stopping potential is independent of the radiation intensity. The maximum kinetic energy of the photoeletrons is related to the stopping potential through the relationship, KE max = e - V s The maximum kinetic energy for liberated photoelectrons is KE max = hf - φ Where φ is called the work function of the metal, the minimum energy with which an electron is bound in the metal, can be found in a table Diffraction of X-rays by crystals: the condition for constructive interference is 2dsin Θ = m λ , where m = 1, 2, 3…. The change in wavelength between a scattered xray and an incident ray is called the Compton shift: ∆λ = λ - λ 0 = h/m e c x (1-cos θ ) where m e is the mass of the electron and Θ is the angle between the directions of scattered and incident photons, The quantity h/m e c is called the Compton wavelength and has a value of 0.00243 nm. The Compton shift depends on the scattering angle and not the wavelength. Energy of a photon: E = hf = hc/ λ Momentum of a photon: p = E/c = hc/c λ = h/ λ So, the de Broglie wavelength of a particle is λ = h/p = h / mv Further, frequency of waves is f = E/h The uncertainty principle: If a measurement of the position of
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Given a linked list and two positions ‘m’ and ‘n’. The task is to rotate the sublist from position m to n, to the right… Read More
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Home / Length Conversion / Convert Kiloyard to Ken
# Convert Kiloyard to Ken
Please provide values below to convert kiloyard [kyd] to ken, or vice versa.
From: kiloyard To: ken
### Kiloyard to Ken Conversion Table
Kiloyard [kyd]Ken
0.01 kyd4.3165467626 ken
0.1 kyd43.1654676259 ken
1 kyd431.654676259 ken
2 kyd863.309352518 ken
3 kyd1294.964028777 ken
5 kyd2158.273381295 ken
10 kyd4316.5467625899 ken
20 kyd8633.0935251799 ken
50 kyd21582.73381295 ken
100 kyd43165.467625899 ken
1000 kyd431654.67625899 ken
### How to Convert Kiloyard to Ken
1 kyd = 431.654676259 ken
1 ken = 0.0023166667 kyd
Example: convert 15 kyd to ken:
15 kyd = 15 × 431.654676259 ken = 6474.8201438849 ken
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Posted: December 16th, 2022
# The following data summarize the results from an independent-measures
1. The following data summarize the results from an independent-measures study
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comparing three treatment conditions.
Treatment
I II III
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0 2 4 N=18
0 3 2 G=36
0 1 4 ∑X2=114
3 3 3
0 2 4
0 1 4
M=0.5 M=2 M=3.5
T=3 T=12 T=21
SS=7.5 SS=4 SS=3.5
a. Use an ANOVA with α=.05 to determine whether there are any
significant differences among the three treatment means.
b. Calculate η2 to measure the effect size for this study.
2. The following data summarize the results from an independent-measures study
comparing three treatment conditions.
Treatment
I II II
4 1 0 N=12
6 4 2 G=36
3 5 0 ∑X2=164
7 2 2
M=5 M=3 M=1
T=20 T=12 T=4
SS=10 SS=10 SS=4
a. Calculate the sample variances for each of the three samples.
b. Use an ANOVA with α=.05 to determine whether there are any significant differences among the three treatment means.
3. The following values are from an independent-measures study comparing three treatment conditions.
Treatment
I II III
n=10 n=10 n=10
SS=63 SS=66 SS=87
a. Compute the variance for each sample.
b. Compute MSwithin which would be the denominator of the F-ratio for an ANOVA.
Because the samples are all the same size, you should find that MSwithin is equal to the average of the three sample variances.
4. The following summary table presents the results from an ANOVA comparing four treatment conditions with n=12 participants in each condition. Complete all missing values. (Hint: Start with the df column.)
Source SS df MS
Between Treatments _____ _____ _____ F = 2.50
Within Treatments 88 _____ _____
Total _____ _____
5. One possible explanation for why some birds migrate and others maintain year round residency in a single location is intelligence. Specifically, birds with small brains, relative to their body size, are simply not smart enough to find food during the winter and must migrate to warmer climates where food is easily available (Sol, Lefebvre, & Rodriguez- Teijeiro, 2005). Birds with bigger brains, on the other hand, are more creative and can find food even when the weather turns harsh. Following are hypothetical data similar to the actual results. The numbers represent relative brain size for the individual birds in each sample.
Non-Migrating Short-Distance Migrants Long Distance Migrants
18 6 4 N=18
13 11 9 G=180
19 7 5 ∑X2=2150
12 9 6
16 8 5
12 13 7
M=15 M=9 M=6
T=90 T=54 T=36
SS=48 SS=34 SS=16
a. Use an ANOVA with α=.05 to determine whether there are any significant mean differences among the three groups of birds.
b. Compute η2, the percentage of variance explained by the group differences, for these data.
c. Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size.
d. Use the Tukey HSD posttest to determine which groups are significantly different.
6. A published report of a repeated-measures research study includes the following description of the statistical analysis. “The results show significant differences among the treatment conditions, F(2,20) = 5.00, p< .05.”
a. How many treatment conditions were compared in the study?
b. How many individuals participated in the study?
7. A recent study examined how applicants with a facial blemish such as a scar or birthmark fared in job interviews (Madera & Hebl, 2011). The results indicate that interviewers recalled less information and gave lower ratings to applicants with a blemish. In a similar study, participants conducted computer-simulated interviews with a series of applicants including one with a facial scar and one with a facial birthmark. The following data represent the ratings given to each applicant.
Applicant
Participant Scar Birthmark No Blemish Person Totals
A 1 1 4 P = 6
B 3 4 8 P = 15 N = 15
C 0 2 7 P = 9 G = 45
D 0 0 6 P = 6 ∑X2 = 231
E 1 3 5 P = 9
M=1 M=2 M=6
T=5 T=10 T=30
SS=6 SS=10 SS=10
a. Use a repeated-measures ANOVA with α=.05 to determine whether there are significant mean differences among the three conditions.
b. Compute η2, the percentage of variance accounted for by the mean differences, to measure the size of the treatment effects.
c. Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size.
8. The following data are from an experiment comparing three different treatment conditions:
A B C
0 1 2 N=15
2 5 5 ∑X2=354
1 2 6
5 4 9
2 8 8
T=10 T=20 T=30
SS=14 SS=30 SS=30
a. If the experiment uses an independent-measures design, can the researcher conclude that the treatments are significantly different? Test at the .05 level of significance.
b. If the experiment is done with a repeated-measures design, should the researcher conclude that the treatments are significantly different? Set alpha at .05 again.
9. The following summary table presents the results from a repeated-measures ANOVA comparing three treatment conditions with a sample of n=12 subjects. Fill in the missing values in the table. (Hint: Start with the df values.)
Source SS df MS
Between treatments _____ _____ 10 F = _____
Within treatments _____ _____
Between subjects _____ _____
Error 44 _____ _____
Total 106 _____
10. The following matrix presents the results from an independent-measures, two-factor study with a sample of n=10 participants in each treatment condition. Note that one treatment mean is missing. Factor B
B1 B2
A1
Factor A
A2
a. What value for the missing mean would result in no main effect for factor A?
b. What value for the missing mean would result in no main effect for factor B?
c. What value for the missing mean would result in no interaction?
Extra Credit (+1)
The following table summarizes the results from a two-factor study with 2 levels of factor A and 3 levels of factor B using a separate sample of n=11 participants in each treatment condition. Fill in the missing values. (Hint: Start with the df values.)
Source SS df MS
Between Treatments 124 ____
Factor A _____ _____ _____ F = 10
Factor B _____ _____ _____ F = _____
A X B Interaction 20 _____ _____ F = _____
Within Treatments _____ _____ 4
Total _____ _____
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# Index an arbitrary array in NDSolve to simulate white noise?
Posted 7 months ago
1067 Views
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7 Replies
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3 Total Likes
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I'm trying to introduce a randomized element in a simulation, using NDSolve to do the integration. The random element is supposed to simulate a random thermal force. The way I've tried to implement this is by creating an array of random values and then trying to select values from the array during each time step. The following is a snippet of my current code. (*Initialising some time values that will allow me to select particular values from my random array*) Tfinal = 10; SampleFreq = 2500; NoT = Tfinal*SampleFreq + 1; dt = Tfinal/(NoT - 1); (*Array of random values and the function I'll be using in NDSolve*) WN = RandomVariate[NormalDistribution[0, 1], NoT]; WNoise[t_] := Indexed[WN, Floor[Floor[t/dt]] + 1] (*NDSolve function using white noise function*) sol = x[t] /. NDSolve[{D[x[t], t, t] == -(w^2) x[t] + WNoise[t], x[0] == 0, x'[0] == 2 }, x, {t, 0, 10}] The function above (without white noise) is the classic F = -kx equation. I've just tried to introduce a white noise driving force force to it as well. Unfortunately, it doesn't work when the WNoise function is added. The error message I typically get is the following. Indexed::ind: The index 1/2 is not a nonzero integer. NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 0.. I'm just wondering if anyone has faced a similar problem and if this method of introducing white noise is possible or if I'm just being silly with this implementation.Thank you for your help.
7 Replies
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Posted 7 months ago
Sandip,You can't mix a digitized noise signal with a continuous integration that way. The derivatives are evaluated along continuous time points and are not on fixed intervals. You have several options. You can use WhenEvent to trigger an event at each sample point and add in the noise or you can use Interpolation to make your noise continuous. I'll show the later. You also need to decide the frequency of the noise (SampleFreq) because a very high rate will significantly slow down the integration. Tfinal = 10; SampleFreq = 2500; NoT = Tfinal*SampleFreq + 1; dt = Tfinal/(NoT - 1); (*NDSolve function using white noise function*) WN = RandomVariate[NormalDistribution[0, 1], NoT]; noise = Interpolation[Transpose[{Range[0, 10, 1/SampleFreq], WN}]]; sol = x[t] /. NDSolve[{D[x[t], t, t] == -(w^2) x[t] + 100*noise[t], x[0] == 0, x'[0] == 2}, x, {t, 0, 10}]; Plot[sol[[1]], {t, 0, 10}] Regards,Neil
Posted 7 months ago
Hi, Neil.Thank you so much! Your solution is exactly what I was looking for. However, I am quite new to Mathematica and I was wondering if you could explain to me what exactly the function Interpolation[Transpose[{Range[0, 10, 1/SampleFreq], WN}]] is doing. Naively, to me, it looks like it's taking the WN array and interpolating between the different values, with a frequency of SampleFreq from t=0 to t=10. However, I don't quite understand the purpose of the Transpose function. Thank you again for your help, Neil.Best Regards,Sandip
Posted 7 months ago
Other responses are probably a better way to go but I want to mention that this can be made to work with very simple modifications.(1) Give w a value.(2) Define WNoise so that it only evaluates for explicit numeric values of t. WNoise[t_?NumberQ] := Indexed[WN, Floor[t/dt] + 1] (3) Also it would be good to maintain consistency with time parameters, so run NDSolve from 0 to Tfinal.A possible advantage to this approach is that it does not try to create a differentiable function from the random impulses. Which is no guarantee NDSolve won't treat it as differentiable using numeric finite differences (though I do not think it will do that).
Posted 7 months ago
Sandip,Glad I could help. You created a list of random samples and you wanted them at SampleFreq spacing to represent noise with a certain frequency content. To do this I had to create a continuous time function with your noise. Interpolation[] takes a list of pairs in the form of {time, value}. I used a "trick" to create the list of pairs: I created the time base as a vector (list) of times with Range[0,10,1/SampleFreq] -- {0, 1/2500, 1/1250,...} I made a matrix out of it by putting the times and values in a list: {{0, 1/2500, 1/1250,...}, {.111, .222, .333, ...}} This is a matrix with two rows and many columns, However the format is wrong for Interpolate -- I need pairs of points which consist of many rows of two Columns. I use Transpose[] on the matrix to make it have the correct format. as an example: In[1]:= lst = {{1, 2, 3, 4, 5}, {a, b, c, d, e}} Out[1]= {{1, 2, 3, 4, 5}, {a, b, c, d, e}} In[2]:= Transpose[lst] Out[2]= {{1, a}, {2, b}, {3, c}, {4, d}, {5, e}} In matrixform:I used Interpolation to create a function of time that the integrator can use as any other continuous function in MMA.Now you may want to consider some options to Interpolation because the default is 3rd order interpolation. If you are modeling a digital system with noise you may want to use InterpolationOrder->0 to get a sample-and-hold version.The Transpose[] trick is worth remembering because it is frequent that you have lists in the wrong format and need to rearrange things. I call it a "trick" because most people do not think of matrix operations when handling lists but they can be useful.Regards,Neil
Posted 7 months ago
Try searching http://reference.wolfram.com/language/ref/ItoProcess.html (the doc. page for ItoProcess) for "white" (or "white noise"). You'll should find an example very similar to what you're trying to do.
introduce a randomized element in a simulation base simulation is of F = -kx, with heat introducing a random element it's vague to me what you are simulating. i see frequency and unsure what it is the frequency of (thus, how heat could physically effect it). is it a spring? is random heat lengthening/softening the spring but random cooling also occurs? (but that situation doesn't make sense to simulate)but if the randomness were "average random", then overall there would be no change in derivatives or integralswithout knowing what you are trying to simulate suggesting answers doesn't make sense to mesince both eq. above are solved in many books, there would be no need to solve them (you could look them up). but mm can do both forms I'm sure Neil Singer (1st responder) knows better than me as to answering. I see your equation as form (t->x, using y,x as typical, and Q(x) representing the random (equation of in x if we call it that, and what kind matters)): y' + c y = Q(x) `that's certainly solvable if entered into DSolve correctly. and you can solve for Yp for each Q(x) or for fun entertain a family of Q(x) as Yp.however for y = -kx, the "standard" eq. (time, motion, position) to use is:y'' + w0^2 y = F sin(wt + B) (* forced undamped motion is the example, w is a constant *)which is stolen from p. 365 of Ord. Diff. Eq. (tenenbaum, pollard). you might then wonder at the solutions of that if B is random. but it would be best to not do that and speculate for that particular case that it would result in random phase change(if you didn't know, Y(x)=Yc+Yp, where Yp =0 if Q(x)=0, that is solutions for Q(x) are simply added to the general solution Yc which is when Q(x)=0)MAYBE this will help answer the question. for y''+w0^2y=F sin(wt+B), if w0 is close to w, wild surges occur, otherwise we may see damping (a ring, peters out), or sin some situations a gradual rise or fall in a range (not a full complete sine wave of motion).SUMMARY of MAYBE: if you take the opposite of what Neil Singer said (force the randomness to have a frequency) and do NOT force a frequency, then you can say nothing of the result. it could be zero by chance, increase or decrease to zero, it could decrease to zero then revive itself.
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### Math and Meter
What is meter
Subject
Topic
Level
Primary
5
How do you solve Find the cost of 7.5meters of cloth at \$67.459meter
Subject
Level
Primary
6
The length and breadth of a rectangular field are 64m and 16m respectively if is exchanged field of the same area find the side of the square field
Subject
Level
Junior High
7
The sides of a rectangular park are in the ratio 4:3. If its area is 1728 meters squared, find the cost of fencing it at ₹30 per metre.
Subject
Level
Middle School
7
Subject
Level
Highschool
7
Subject
Level
Primary
7
2questions
Subject
Level
Middle School
8
How to solve question 20??
Level
Primary
4
Subject
Level
Middle School
5
Subject
Level
Primary
4
A wire in the shape of a square of side 11m is rebent intothe shape of a circle.Find the area of the circle?
Subject
Level
Junior High
7
The length of a lawn is 13.5m and breadth is 10m. Find the cost of the paving the lawn at the rate of Rs50 per square metres.
Subject
Level
Middle School
6
How to find the greatest area
Subject
Level
Middle School
6
Help
Subject
Level
Middle School
4
You have a bike with tires that have a radius of 40cm how far will the bike travel if you make 450 rotations
Subject
Level
Middle School
7
I need help for math
Subject
Level
Middle School
7
A piece of wire is in the shape of an equilateral triangle whose side measures 4.4 cm each this wire is rebent to form a circular ring what is the area of the ring (π=22/7)
Subject
Level
Primary
7
A trapezoid has a base of 9 inches and 7 inches and a height of of 5 inches what is the area
Subject
Level
Primary
4
(Area and perimeter) a verandah of width 2.25m is constructed all along outside a room which is 5.5m long and 4m wide. Find :
(a) the area of verandah
(b) the cost of cementing the floor of the verandah at the rate of \$200 per m square.
Subject
Level
Highschool
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## Results for: Stoller-robert-j
In Personal Finance
# What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow (MORE)
In Acronyms & Abbreviations
# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo (MORE)
In Coins and Paper Money
# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
In Math and Arithmetic
# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
In Volume
# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
In Numerical Analysis and Simulation
# What is the answer for 5c equals -75?
The 'answer' is the number that 'c' must be, if 5c is really the same as -75. In order to find out what number that is, you could use 'algebra'. First, write the equatio (MORE)
In iPhone 5
# How many pixels does the iPhone 5c have?
The iPhone 5c is 640 x 1136 pixels. That is about 326 pixels persquare inch (ppi).
In Temperature
# What is minus 5c in Fahrenheit?
(-5) degrees Celsius = 23 degrees Fahrenheit. Formula: [°F] = [°C] à 9 â 5 + 32
In iPhone 5
# How many inches is a iPhone 5c?
The screen is 4" big. The height is 4.9", width is 2.33" and thedepth is 0.35"
In iPhone 5
# How much does an iPhone 5c weigh?
The iPhone 5c weighs 4.65 ounches. It is heavier than the iPhone 5and 5s which weight 3.95.
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# Detecting Data Leakage
Hector Garcia-Molina
hector@cs.stanford.edu
Leakage Problem
Name: Sarah
Sex: Female …. Name: Mark
Sex: Male
…. Jeremy Sarah App. U1 App. U2 Mark
Other Sources e.g. Sarah’s Network
Kathryn
Stanford Infolab
2
7 – Pr{U2 leaked data} = 0.2 • Distribution Strategies Stanford Infolab 3 .Outline • Problem Description • Guilt Models – Pr{U1 leaked data} = 0.
• Problem Description • Guilt Models • Distribution Strategies Stanford Infolab 4 .
Rn Ri: Set of people’s profiles who have added the application Ui Leaker S Set of leaked profiles Stanford Infolab 5 . ….Problem Entities Entity Distributor Facebook Dataset T Set of all Facebook profiles Agents Facebook Apps U1. Un R1. ….
Agents’ Data Requests • Sample – 100 profiles of Stanford people • Explicit – All people who added application (example we used so far) – All Stanford profiles Stanford Infolab 6 .
• Problem Description • Guilt Models • Distribution Strategies Stanford Infolab 7 .
given the leaked set of profiles S Stanford Infolab Other Sources e. Sarah’s Network 8 .Guilt Models (1/3) p: posterior probability that a leaked profile comes from other sources p p Guilty Agent: Agent who leaks at least one profile Pr{Gi|S}: probability that agent Ui is guilty.g.
Guilt Models (2/3) Agents leak each of their data items independently p2 Agents leak all their data items OR nothing p(1-p) (1-p)p or or (1-p)2 or Stanford Infolab 9 .
Guilt Models (3/3) Independently NOT Independently Pr{G2} Pr{G2} Pr{G1} Pr{G1} Stanford Infolab 10 .
• Problem Description • Guilt Models • Distribution Strategies Stanford Infolab 11 .
The Distributor’s Objective (1/2) R1 R2 U1 U2 R3 S (leaked) R1 R3 R4 U3 U4 Stanford Infolab Pr{G1|S}>>Pr{G2|S} Pr{G1|S}>> Pr{G4|S} 12 .
j 1..The Distributor’s Objective (2/2) • To achieve his objective the distributor has to distribute sets Ri. Rn that minimize i 1 Ri R R j i i j ... n • Intuition: Minimized data sharing among agents makes leaked data reveal the guilty agents Stanford Infolab 13 .. i. ….
Sarah and Mark • There are 4 agents: – U1.Distribution Strategies – Sample (1/4) • Set T has four profiles: – Kathryn. U3 and U4 • Each agent requests a sample of any 2 profiles of T for a market survey Stanford Infolab 14 . Jeremy. U2.
Distribution Strategies – Sample (2/4) Poor Minimize Ri R j i j U1 U2 U3 U4 U1 U2 U3 U4 Stanford Infolab 15 .
Distribution Strategies – Sample (3/4) • Optimal Distribution U1 U2 U3 U4 i • Avoid full overlaps and minimize R R j i i 1 Ri j Stanford Infolab 16 .
Distribution Strategies – Sample (4/4) Stanford Infolab 17 .
g..Distribution Strategies Sample Data Requests • The distributor has the freedom to select the data items to provide the agents with • General Idea: – Provide agents with as much disjoint sets of data as possible Explicit Data Requests • The distributor must provide agents with the data they request • General Idea: – Add fake data to the distributed ones to minimize overlap of distributed data • Problem: There are cases where the distributed data must overlap E. |Ri|+…+|Rn|>|T| • Problem: Agents can collude and identify fake data • NOT COVERED in this talk Stanford Infolab 18 .
Conclusions • Data Leakage • Modeled as maximum likelihood problem • Data distribution strategies that help identify the guilty agents Stanford Infolab 19 .
Thank You! .
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# How do I think about solving this sort of problem without having to count intersections?
I have asked this sort of question before, and I have a new similar question here. The latter question got me thinking about this specific question.
When throwing 5d20 (rolling five twenty sided polyhedral dice), what is the probability of three or more dice having a result of '5' or less, of the three, two must have a result of '4' or less? (The remaining two dice of the five I don't care about)
(Example: Roll 5d20, with a result of (15,4,7,2,1) on the (1st, 2nd, 3rd, 4th, and 5th) die. This would be considered a success, because the '4' is five or less, and the '2' & '1' are four or less.)
I can do some simple calculations like
• What are all the permutations of 5d20? $20\times 20\times 20\times 20\times 20 = 3.2$ million
• What are the ways of arranging '5','4','4', x , x . (where x is not '4', or '5')
1) 544xx, 54x4x, 54xx4, 5x44x, 5x4x4, 5xx44, (6 ways with '5' in 1st position)
2) 454xx, 45x4x, 45xx4, x544x, x54x4, x5x44, (6 ways with '5' in 2nd position)
3) ... (6 ways with '5' in 3rd position)
For a total of 30 different ways of arranging the 544xx combination. If the x can represent any dice face other than '5' or '4' (to avoid counting duplicates), that gives me $30\times 18\times 18 = 9720$ successes so far. But, there are so many more successes that I haven't counted. I have to worry about the (5,4,4,4,x) throws which I didn't count before because I didn't want to worry about duplication, and the (5,4,4,4,4) throws which is admittedly small (only 5 permutations of the 3.2 million). Then there are all the other permutations I need to count.
1) (5,4,3,x,x) throws, (5,4,2,x,x) throws, (5,4,1,x,x) throws,
2) (5,3,3,x,x) throws, (5,3,2,x,x) throws, (5,3,1,x,x) throws,
3) ...
Each time I have to worry about the intersection of the permutations (i.e. In the above examples, I need to worry about double counting results where the 'x' represents everything except the dice I am counting all the permutations of for a certain the combinations of dice.), and each time I have to worry about missing counting a success. This can be alleviated somewhat by counting the failures, and making sure that the successes+failures adds up to 3.2 million total permutations that I know exist, but this is ripe for mistakes.
• I can brute force it with a simple computer program, (3.2 million permutations is only a few seconds). This is a rather easy solution for me because I am a programmer, and not a mathematician
Note: The brute force solution in pseudo code is
iterate over die_1 from 1 to 20
iterate over die_2 from 1 to 20
iterate over die_3 from 1 to 20
iterate over die_4 from 1 to 20
iterate over die_5 from 1 to 20 {
bubblesort die_1 through die_5 (biggest die is die_5)
if die_3 is 5 or less, and die_2 is 4 or less, success_count increment
else failure_count increment
}
The result of the above is 300,704 successes (one '5' or less, and two additional dice '4' or less), out of 3.2 million.
but, what I really want to know is...
• How should I think about solving this sort of problem so that I can either easily identify intersections between sets, or avoid double counting the intersections?
• Is this even possible, or should I just brute force my solutions?
Note: I just noticed that I could solve this question easier by thinking about it as a tree, and a single d20 being rolled. For each roll, I am either closer to obtaining a success, or not.
1st Roll: 1/20 '5'
4/20 '4' or less
15/20 no successes
2nd Roll: from '5' : 4/20 '4'or less
from '5' : 16/20 no additional successes.
from '4'< : 1/20 '5'
from '4'< : 4/20 '4' or less
from '4'< : 1/20 no additional successes.
from 'no success' : 1/20 '5'
from 'no success' : 4/20 '4' or less.
from 'no success' : 15/20 no successes
3rd Roll: keep applying odds at each step.
If complete success (found 3 dice that fit the criteria), multiply result by 20 to the power of X, where X is the number of remaining throws. This tree can also be used to find the failures at each step. (Example: if the 1st & 2nd & 3rd die rolls were all greater than 6, then that part of the tree is a failure. Multiply the number of cases (15*15*15) times the remaining die roll (20*20) to get the number of failures for that part of the tree. All branches of the tree (successes and failures) will total the number of permutations (3.2 million).
-
Note that * is a special character in this site, which triggers italics or boldface; also, you were using * to mean two different things in the same post: multiplication in some places, and a placeholder in others. It is very bad form, and very confusing, to use the same symbol to mean two different things in the same argument. I've edited to replace these problems. Please double-check to make sure I didn't change the meaning by mistake. – Arturo Magidin Jan 30 '12 at 16:53
I'm not clear on your "three dice with 5 or less, two dice with 4 or less". Do you simply mean "at least one five and two fours"? – Arturo Magidin Jan 30 '12 at 16:55
@ArturoMagidin, it must have been a typo. What I meant was at least one die with 5 or less, and at least two other/different dice with 4 or less. This is of course different than at least one '5', and two '4's. At least one '5' & two '4's doesn't count a result of '3', '2', '2', 'X', 'X' which would be a success in my example. – user1873 Jan 31 '12 at 2:44
I guess my confusion comes from the fact that a roll in which at least two dice are 4 or less already guarantees one dice that is five or less. I guess what you really mean is that at least three dice are not 6, and of those three, at most one is equal to 5. Yes? – Arturo Magidin Jan 31 '12 at 3:35
@ArturoMagidin, yes. That isn't how I would word it, but you do have it correct. At least three dice are not "6 or more" (there exists at most two dice that have values of 6 or more), and of the three that are not 6 or more, at most one of them may have a value of 5. Is there some easier way of stating this? Each time I rewrite it, it seems less intuitive. – user1873 Jan 31 '12 at 4:29
## 2 Answers
Let $X$ be the number of rolls of 4 or less, $Y$ the number of rolls of exactly 5. You want the probability that $X+Y \ge 3$ and $X \ge 2$. This is $\sum_{x=2}^5 \sum_{y=\min(3-x,0)}^{5-x} P(X=x, Y=y) = P(X=2,Y=1) + P(X=2,Y=2) + P(X=2,Y=3) + P(X=3,Y=0)+P(X=3,Y=1)+P(X=3,Y=2)+P(X=4,Y=0)+P(X=4,Y=1)+P(X=5,Y=0$ where (since on a single die you have probability $4/20$ of getting 4 or less and $1/20$ of getting 5) $$P(X=x, Y=y) = \frac{5!}{x! y! (5-x-y)!} (4/20)^x (1/20)^y (15/20)^{5-x-y}$$
-
although this is an answer to my example problem (is it correct? do you get 300,704), I was looking at a more general way at looking at this type of problem. When I thought about the question again in the terms of a tree, and how close I was to achieving my objective, it reminded me of a markov chain. My other die question (small straight 6d6) linked above didn't use the same method to achieve the solution. Brute force was the easiest solution for many of the people who provided answers. – user1873 Jan 31 '12 at 4:35
This is rather unimaginative, but you could do it in the straightforward way:
If you want "at least three rolls less than or equal to 5 and at least two less than or equal to 4":
First, split this up into the disjoint events
$\ \ A$: exactly three rolls less than or equal to 5 and at least two less than or equal to 4
$\ \ B$: exactly four rolls less than or equal to 5 and at least two less than or equal to 4
$\ \ C$: exactly five rolls less than or equal to 5 and at least two less than or equal to 4
Then split $A$ into the disjoint events :
$\ \ A_1$: exactly one roll equal to five and exactly two rolls less than 5
$\ \ A_2$: exactly three rolls less than 5
Split $B$ into the disjoint events :
$\ \ B_1$: exactly one roll equal to five and exactly three rolls less than 5
$\ \ B_2$: exactly two rolls equal to five and exactly two rolls less than 5
$\ \ B_3$: exactly four rolls less than 5
I'll leave it to you to split up $C$.
-
I don't think you got my meaning in the question. This might because you don't have a role playing background. I will update the question so that you can better understand it. 5d20 means "roll five 20-sided polyhedral dice". What I was interested was how to think about the problem in a general sense so that I can apply this to other dice questions (instead of brute forcing every example). – user1873 Jan 31 '12 at 2:47
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# Physical Science
posted by Anonymous
How is friction involved in sledding and skiing?
1. bobpursley
How do you turn, or stop?
2. Anonymous
hifwo
## Similar Questions
1. ### physical science
gimme a side of what physical science is. I know there's no shortcuts but it means alot of ev'rything I gotta know. Sho' me. I have little understanding of what you wrote. When one tries to communicate in non-standard English, the …
2. ### Intro to Physical Science
Define physical science Andrew: I went to www.dictionary.com and typed in physical science. The following is what it returned.I hope this helps. physical science 1. any of the natural sciences dealing with inanimate matter or …
3. ### Science
How is friction involved in Volleyball? Hand touches volleyball maybe.... i know that but what type would that be?
4. ### English
1. I often go skiing with my friends on weekends. 2. I often go skiing with my friends weekends. 3. I often go skiing with my friends on the weekends. 4. I often go skiing with my friends on the weekend. -------------------------- …
5. ### English
1. I often go skiing with my friends on weekends. 2. I often go skiing with my friends weekends. 3. I often go skiing with my friends on the weekends. 4. I often go skiing with my friends on the weekends. Are they all grammatical?
Need help on this question. What happened last?
7. ### Math
104 students responses 21 liked sledding as their favorite winter activity if 500 people had. Responded, how many would have been expected to list sledding as their favorite winter activity?
8. ### physical science
what causes friction?
9. ### Science
john is on a skiing holiday. he starts from rest on a long gentle incline of 6° and the coefficient of kinetic friction between the skies and the snow is 0,06. the weight of john and his skies is 800 n. Traveling 100m
10. ### Science homework
john is on a skiing holiday. he starts from rest on a long gentle incline of 6° and the coefficient of kinetic friction between the skies and the snow is 0,06. the weight of john and his skies is 800 n., traveling 100m
More Similar Questions
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# Blueprint Loops, Objects, and Logic
This is more of a “is there a better way” sort of question. I find myself struggling with arrays, loops, and object communication in Unreal Engine 4. Specifically adding elements to an array and pulling information out via a loop. To help my brain, I watched Epic’s live stream on Blueprint communication but it isn’t my mental go-to.
Here’s a scenario: Suppose you have some number of puzzle cubes in a scene which need to be toggled in a specific order; get the wrong order, they all reset. To check this puzzle in game, we’ll need to know the number of cubes and the state of each cube.
In C, I would do it like this:
``````
// loop through all cubes
for(n = 0; n < numCubes; n++)
{
// add cubes to count if toggled on
if(cube[n].isOn && !cube[n].isCounted)
{
cube[n].isCounted = true;
cubesCounted++;
}
// remove cubes from count if toggled off
else if(!cubes[n].isOn && cube[n].isCounted)
{
cube[n].isCounted = false;
cubesCounted--;
}
}
// check puzzle status
if(numCubes == cubesCounted)
puzzleFinished = true;
eles
puzzleFinished = false;
``````
In Unreal, however, I find myself tripping over Blueprint code – like sometimes, I just don’t get how these bubbles communicate. Below is a solution I came up with, but I feel I’m going about it the wrong way. Thoughts?
Move complicated logic to c++.
or
UE4 Blueprints From Hell
• You don’t need the Get node in a ForEachLoop, just use the Array Element
• You should never use GetAllActorsOfClass in a Blueprint Tick function
• Make the logic Event based so it only has to check when a variable changes
• As already suggested move this to C++ as it would only need a few lines of code compared to 50 nodes with spaghetti
Also, the ++ & – nodes set by reference - no need to Set it again. 2 nodes down, 48 to go…
Huh…that was the first thing I tried but it wouldn’t get the element’s variable; using “get copy” did which is why I used it. I just tried it again though and it works now. Thanks.
Oh, it works that way? Good to know. Thank you.
Agreed but I’m trying to…learn this because I have to teach it and my students go pale at the sight of code.
Is there no god?
On the contrary
A friend of mine suggested this:
``````
// function check for puzzle
int PuzzleFinished()
{
// loop through the cubes;
for(n = 0; numCubes; n++)
{
if(!cube[n].isOn)
return false;
}
// if loop finished
return true;
}
``````
And I was like – of course! So the new blueprint looks like this:
So, as GarnarP57 said, it’s best to run it off an event. The function runs only if we need to know the variable and the loop runs through each of the elements. If at any point the loop finds an element that isn’t on, it breaks the loop and returns false. If the loop finishes, it then returns true since they’ll all be on. Thus far, it works with all of my tests.
Tip:
In that example you actually don’t need a “loop with break” you can simple make a Return false node if IsOn=false. Then the function (and the loop) will end early. If it doesn’t end early it will reach “Completed” and you can return true.
You’re right, because the return would exit the loop upon the first instance of !isOn. Thanks. I’m still trying to wrap my brain around bubble scope. To me, those returns are outside the loop body but I guess not lol.
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A322145 Number of permutations of 9 copies of 1..n with no element equal to another within a distance of 1. 2
1, 0, 2, 17236524, 2511603532825176, 2829059722872229922701920, 17498057808683351584656839871450000, 459422439054082909311010463927575656038701920, 42176005899746902650961357272521722186133207293858938240 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS Seiichi Manyama, Table of n, a(n) for n = 0..64 FORMULA a(n) = n! * A321669(n). a(n) = Integral_{0..infinity} (Sum_{k=1..9} (-1)^(9-k) * binomial(8, 9-k) * x^k/k!)^n * exp(-x) dx. CROSSREFS Row 9 of A322093. Cf. A321669, A322127, A322128. Sequence in context: A326615 A259407 A157992 * A301810 A296627 A247214 Adjacent sequences: A322142 A322143 A322144 * A322146 A322147 A322148 KEYWORD nonn AUTHOR Seiichi Manyama, Nov 28 2018 STATUS approved
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# Column Matrix
## Definition
A matrix whose all elements are arranged in a column is called a Column matrix.
Column matrix is a type of matrix and it is also called as a column vector. All elements in this type of matrix are arranged in different rows but only in one column.
$M$ is a column matrix in general form and it is known as a column matrix of order $m \times 1$.
$M = {\begin{bmatrix} e_{11}\\ e_{21}\\ e_{31}\\ \vdots\\ e_{n1} \end{bmatrix}}_{\displaystyle m \times 1}$
The column matrix can be expressed in simple form.
$M = {\begin{bmatrix} e_{\displaystyle ij} \end{bmatrix}}_{\displaystyle m \times n}$
Each element in this matrix is arranged in a row but in only one column. Therefore, the column $j = 1$. Therefore, the number of columns is one. The general form column matrix can be expressed as follows.
$M = {\begin{bmatrix} e_{\displaystyle i1} \end{bmatrix}}_{\displaystyle m \times 1}$
### Example
Observe the following examples to understand how elements are arranged in column matrices.
$(1)\,\,\,\,$ $A = \begin{bmatrix} 7 \end{bmatrix}$
$A$ is a column matrix of order $1 \times 1$. In this column matrix, the only one element is displayed in one row and one column.
$(2)\,\,\,\,$ $B = \begin{bmatrix} -1\\ 4 \end{bmatrix}$
$B$ is a column matrix of order $2 \times 1$ and in this matrix, the two elements are arranged in two rows and one column.
$(3)\,\,\,\,$ $C = \begin{bmatrix} 6\\ 0\\ 9 \end{bmatrix}$
$C$ is a column matrix of order $3 \times 1$. The three elements are arranged in the matrix in three rows and one column.
$(4)\,\,\,\,$ $D = \begin{bmatrix} -5\\ 8\\ 2\\ 3 \end{bmatrix}$
$D$ is a column matrix of order $4 \times 1$. The four elements are arranged in the matrix in four rows and one column.
The column vector can have any number of elements but all the elements are arranged in number of rows but only in one column.
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###### Math Problems
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
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Questions
# A bird in air looks at a fish directly below it inside in a transparent liquid in a tank. If the distance of the fish as estimated by the bird is h1 and that of the bird as estimated by the fish is h2, then the refractive index of liquid is
a
h2h1
b
h1h2
c
h1 + h2h1 - h2
d
h1 - h2h1 + h2
detailed solution
Correct option is A
Let h be the height the bird from the water before and d be the depth of the fish from the water surfaceh1 = h +dμ , h2 = d + μh⇒h2h1=d+μh(d+μh)/μ=μ
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# How to Create a Correlation Matrix in SAS (With Example)
A correlation matrix is a square table that shows the correlation coefficients between variables in a dataset.
It offers a quick way to understand the strength of the linear relationships that exist between variables in a dataset.
You can use the PROC CORR statement in SAS to create a correlation matrix for a given dataset:
```/*create correlation matrix using all numeric variables in my_data*/
proc corr data=my_data;
run;```
By default, this will create a matrix that displays the correlation coefficients between all numeric variables in the dataset.
To only include specific variables in the correlation matrix, you can use the VAR statement:
```/*create correlation matrix using only var1, var2 and var3 in my_data*/
proc corr data=my_data;
var var1, var2, var3;
run;```
The following example shows how to create a correlation matrix in SAS.
## Example: Creating a Correlation Matrix in SAS
Suppose we have the following dataset in SAS that contains information about various basketball players:
```/*create dataset*/
data my_data;
input team \$ assists rebounds points;
datalines;
A 4 12 22
A 5 14 24
A 5 13 26
A 6 7 26
B 7 8 29
B 8 8 32
B 8 9 20
B 10 13 14
;
run;
/*view dataset*/
proc print data=my_data; ```
We can use the PROC CORR statement to create a correlation matrix that includes each numeric variable in the dataset by default:
```/*create correlation matrix using all numeric variables in my_data*/
proc corr data=my_data;
run;```
The output displays summary statistics of the numeric variables in the first table along with a correlation matrix.
Note that the “team” variable was not included in the correlation matrix because it was not a numeric variable.
Here is how to interpret the values in the correlation matrix:
(1) The Pearson correlation coefficient (r) between assists and rebounds is -0.24486. The corresponding p-value is 0.5589.
Since r is less than zero, this tells us that there is a negative linear association between these two variables. However, the p-value is not less than .05 so this correlation is not statistically significant.
(2) The Pearson correlation coefficient (r) between assists and points is -0.32957. The corresponding p-value is 0.4253.
There is a negative linear association between these two variables but it is not statistically significant.
(3) The Pearson correlation coefficient (r) between rebounds and points is -0.52209. The corresponding p-value is 0.1844.
There is a negative linear association between these two variables but it is not statistically significant.
Note that we could also use the VAR statement to only include specific numeric variables in the correlation matrix:
```/*create correlation matrix using only assists and rebounds variables*/
proc corr data=my_data;
var assists rebounds;
run;```
Notice that only the assists and rebounds variables were included in this correlation matrix.
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https://www.cfd-online.com/Forums/main/312-what-density-based-method-print.html
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CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- Main CFD Forum (https://www.cfd-online.com/Forums/main/)
- - What is Density-Based method? (https://www.cfd-online.com/Forums/main/312-what-density-based-method.html)
C-H Kuo November 11, 1998 10:20
What is Density-Based method?
This is a great place for post-school learning. I appreciate the large and diversified knowledge from this website.
I have been working on CFD several years, and once a while I hear people talking about pressure-based and density-based solution methods. I really don't know what it is, could not find from literature, and felt shame to ask. I asked several times and did not get it. I think it is an ambiguous term or the person I asked is same as me--don't really know it.
Could someone give a detailed description on this method, and the advantage and disadvantage of it. I will appreciate this, and I belive many out there will also appreciate it.
Hongjun Li November 11, 1998 13:09
Re: What is Density-Based method?
There is realy no explict difinition of 'density-based' or 'pressure based' (or as far as I know). But the two solution methods are quite different. The main difference is how to link the continuity equation to the other equations (momentum and energy) so that the conservations of mass, momentum and energy are satisfied simutanously during the course of solution.
The 'density-based' (DB in short) methods are widely used in compressible N-S equations. No matter for steady-state or unsteady state solutions, the algorithm must be developed from the unsteady compressible N-S equations. The 'links' between the continuty Eq. and the momentum and energy eqs. relies on the density change which is obtained from the continuity eq. Put it in a simle way (in reality, it is not that simple, because everythin is strongly coupled) if the mass is not conservative, there will be a density residual from the continue eq. This density residual will lead to velocity and temperature residuals through the momentum and energy equations. That will updata all flow quantities until conservation arrived. Most(but not all) compressible codes are density-based (MacCormack, Beam-Warming, Jameson, Van Leer, as well as all high order Godunov-type schemes like MUSCL, ENO, TVD etc.) Some advantages of DB schemes are (among others) (1) all equations can be in strongly conservation form, the solution methods are the same for each eq. (2) strong shock waves can be captured. The disadvantage is its poor capability to solve low Mach flows (slow and hard to converge). Most DB code will fail to converge to the RIGHT solution for Mach number below 0.1.
The 'pressure-based' (PB) method used the pressure to link the continuity equation and the others. In this method, the continuity equation must be solved separatly. There are several approaches in this area (please refer to numerical methods for incompressible N-S equations in CFD text books). The PB method was originally for incompressible flows (density term does not appear in continuity equation, even for unsteady flow). A good example of PB code is SIMPLE. But later, this method is extended to solve compressible flows as well, with less ability to handle strong shock waves.
In short statement but not absolute (I have to say that because there are a few people love to make nonsense argument or even personal attack). If you need to due with compressible flow (M=0.3 to very high), you may want to use DB approach. If you need to solve incompressible flows, you may choose PB method. If your problem is more complicated including both low and high speeds, you may try the newly developed 'all-speed method' which is still not very popular yet.
Hope this will help you to get some sense.
Hongjun
jay November 11, 1998 18:06
Re: What is Density-Based method?
In the density based method, the algorithm uses a constitutive relation such as the ideal-gas law to iterate for the pressure. In the pressure based method, no constitutive relation is used; pressure is never even iterated for, only the pressure difference is calculated iteratively (in SIMPLE)
That is the gist of the matter. For an explanation as to why: it has to do w/ the structure of the equations that describe compressible and incompressible flows. One is elliptic, another is parabolic (in basic form; I bet there are people out there who'll jump on me for saying this!)
W.F Ames's book on Numerical Analysis is a good place to start to understand these math details. Someone below has suggested Fletcher for a beginners book. That also is a very good reference to understand this concept.
Aldrin Wong November 18, 1998 01:09
Re: What is Density-Based method?
Where can I find more information regarding the "all-speed" method ? Is there a proper name for it ?
Hongjun Li November 18, 1998 17:11
Re: What is Density-Based method?
I did some work in this area about 4-5 years ago. I need to dig out some papers from my files and come back to you later.
All times are GMT -4. The time now is 05:44.
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# Fall 2019 - MATH 150 D300
## Overview
• #### Course Times + Location:
Sep 3 – Dec 2, 2019: Mon, Wed, Fri, 9:30–10:20 a.m.
Surrey
• #### Exam Times + Location:
Dec 5, 2019
Thu, 3:30–6:30 p.m.
Surrey
• #### Prerequisites:
Pre-Calculus 12 (or equivalent) with a grade of at least B+, or MATH 100 with a grade of at least B-, or achieving a satisfactory grade on the Simon Fraser University Calculus Readiness Test. Students with credit for either MATH 151, 154 or 157 may not take MATH 150 for further credit.
## Description
#### CALENDAR DESCRIPTION:
Designed for students specializing in mathematics, physics, chemistry, computing science and engineering. Topics as for Math 151 with a more extensive review of functions, their properties and their graphs. Recommended for students with no previous knowledge of Calculus. In addition to regularly scheduled lectures, students enrolled in this course are encouraged to come for assistance to the Calculus Workshop (Burnaby), or Math Open Lab (Surrey). Quantitative.
#### COURSE DETAILS:
MATH150 consists of 3 hours of lecture and a 1 hour seminar each week.
Lectures contain both MATH151 and MATH150 students in the same room.
MATH 150 students must register for a 1 hour tutorial/seminar.
Chapter 1 - Functions and Models
1.1 Four ways to represent a function
1.2 Mathematical Models: A Catalogue of Essential functions
1.3 New Functions from Old Functions
1.4 Exponential Functions
1.5 Inverse Functions and Logarithms
Chapter 2 - Limits and Derivatives
2.1 Tangent and Velocity Problems
2.2 Limit of a Function
2.3 Calculating Limits Using the Limit Laws
2.5 Continuity
2.6 Limits at Infinity; Horizontal Asymptotes
2.7 Derivatives and Rates of Change
2.8 The Derivative as a Function
Chapter 3 - Differentiation Rules
3.1 Derivatives of Polynomials and Exponential Functions
3.2 Product and Quotient Rules
3.3 Derivatives of Trigonometric Functions
3.4 The Chain Rule
3.5 Implicit Differentiation
3.6 Derivatives of Logarithmic Functions
3.7 Rates of Change in the Natural and Social Sciences
3.8 Exponential Growth and Decay
3.8 Newton's Law of Cooling
3.9 Related Rates
3.10 Linear Approximations and Differentials
3.11 Hyperbolic Functions (Optional)
Chapter 4 - Applications of Differentiation
4.1 Maximum and Minimum Values
4.2 The Mean Value Theorem
4.3 How Derivatives Affect the Shape of a Graph
4.4 Indeterminate Forms and L'Hospital's Rule
4.5 Summary of Curve Sketching
4.7 Optimization Problems
4.8 Newton's Method
Chapter 10 - Parametric Equations and Polar Coordinates
10.1 Curves Defined by Parametric Equations
10.2 Calculus with Parametric Curves
10.3 Polar Coordinates
• Final Exam 50%
• Midterm 1 15%
• Midterm 2 15%
• Quizzes 8%
• Online Assignments 7%
• Seminar 5%
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## Fractal Landscapes; Variations
These images show the effect of changing the scaling parameter. At each stage in the construction, a random displacement of the midpoints is made according to the formula avg + scale*rand where avg is the average of the adjacent corners, rand is a random number (for these examples, normally distributed about zero with a variance of 1), and scale is a scaling parameter which is proportional to 2d*level where level is k, k-1, k-2, ..., 1 for the first iteration, the second iteration, the third iteration,..., the kth iteration. So the amount of random displacement decreases by a factor of 2d when going to the next iteration (random displacements become smaller as the construction proceeds to smaller and smaller scales). So we expect those landscapes that are made with larger d to have larger extreme variations (due to the first iteration) but relatively smoother surfaces than landscapes made with smaller d (which will have smaller extreme variations but rougher surfaces). You can see that in the examples below. (Imagine walking on these surfaces; for large d you will be going up and down to great heights and depths, but the surfaces of the 'mountains' will be relatively smooth. While for the smaller d surfaces you will not go to great heights or depths, but you will have to walk over very rough surfaces). The fractal dimension of these landscapes is related to the exponent d; larger d correspond to smaller fractal dimensions.
Click on the image for a larger view. Note that the z-scale is not the same for all images.
d=0.1
d=0.3
d=0.5
d=0.7
d=0.9
d=.11
d=.13
d=.15
d=2.0
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https://oeis.org/A303889
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A303889 Number of n X n 0..1 arrays with every element unequal to 1, 2, 3 or 4 king-move adjacent elements, with upper left element zero. 0
0, 7, 54, 1659, 29479, 1541443, 71307109, 5846922441, 668263785713, 103693982275562 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Diagonal of A303896. LINKS EXAMPLE Some solutions for n=5 ..0..0..0..1..1. .0..1..1..1..0. .0..0..1..1..1. .0..1..1..0..1 ..1..0..0..1..0. .0..0..1..1..1. .1..1..1..1..0. .0..1..1..1..0 ..1..0..1..1..0. .0..0..0..0..0. .0..1..0..0..0. .1..1..1..0..0 ..0..0..1..1..1. .1..0..0..0..0. .1..0..0..0..1. .0..1..0..0..0 ..0..1..0..0..1. .0..1..0..1..1. .0..0..1..1..1. .1..0..0..0..1 CROSSREFS Cf. A303896. Sequence in context: A298104 A289865 A182124 * A198149 A203878 A043077 Adjacent sequences: A303886 A303887 A303888 * A303890 A303891 A303892 KEYWORD nonn AUTHOR R. H. Hardin, May 02 2018 STATUS approved
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Last modified June 27 04:16 EDT 2022. Contains 354888 sequences. (Running on oeis4.)
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• ### ball mill initial charge calculationmineralenzoeken
calculate ball mill chargeeitmindiacoin calculation of ball mill charge volume - How to Calculate Charge Volume in Ball or Rod MillMining and The charge volume of a ball or rod mill is expressed as the percentage of the volume within the 24/7 online ball mill initial charge charticchmt2017org
• ### ball mill initial charge calcProductsKefid Machinery
ball mill initial charge calc. 250tph river stone crushing line in Chile. 200tph granite crushing line in Cameroon. 250tph limestone crushing line in Kenya. 250tph granite crushing line in South Africa. 120tph granite crushing line in Zimbabwe. 400tph
• ### Optimum choice of the make-up ball sizes for maximum
It is possible then to incorporate this information into a grinding model and investigate the effect of the make-up ball size on the mill performance with various feed and product specifications. Concha et al. was the first to combine a grinding circuit model with a ball wear model to optimize the make-up ball charge. With an optimization
• ### Ball MillRETSCHpowerful grinding and homogenization
Ball mills are among the most variable and effective tools when it comes to size reduction of hard brittle or fibrous materials. The variety of grinding modes usable volumes and available grinding tool materials make ball mills the perfect match for a vast range of applications.
• ### EFFECTS OF GRINDING MEDIA SHAPES ON BALL MILL
EFFECTS OF GRINDING MEDIA SHAPES ON BALL MILL PERFORMANCE Niyoshaka Nistlaba Stanley Lameck A dissertation submitted to the Faculty of Engineering and The Built Environment University of the Witwatersrand Johannesburg in fulfilment of the requirements for the degree of Master of Science in Engineering Johannesburg October 2005
• ### How can one select ball size in ball milling and how much
How can one select ball size in ball milling and how much material should be taken in mill pot Ball size depends on initial size of charge and ranges from 10mm-150mm dia. I need info
• ### Mill Steel Charge Volume Calculation
We can calculate the steel charge volume of a ball or rod mill and express it as the of the volume within the liners that is filled with grinding media. While the mill is stopped the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charge to the top of the mill.
• ### THE GRINDING CHARGE OF ROTARY MILLS
and the increasing throughput of the mill the charge consists in a mixture of the balls of different sizes. Balls size distribution is charge value to the initial size by adding new grinding media) produces the increasing of the throughput of mill with 15 or even 30
• ### Ball Mill Initial Charge Chart- Jaw crusher ball mill
Ball Mill Initial Charge Chart. The same policy is employed for the makeup charge where adding different ball sizes is done in order to optimize the mill performance the ball mill liners and grinding media employed were made from stainless steel in fig 4 the initial fractions of size Ball mill initial charge
• ### ball mill initial charge chartwonder-parties
ball mill initial charge chart_Ball Mill LoadingDry MillingPaul O. Abbe When charging a ball mill ceramic lined mill pebble mill jar mill or laboratory . simply load the mill with the total amount of
• ### Ball Mill Critical Speed Working PrincipleYouTube
Jun 20 2015 · Learn about Ball Mill Critical Speed and its effect on inner charge movements. The effect of Ball Mill RPM speed going from sub-critical to super-critical helps understand the Ball Mill Working
• ### ball mill initial charge chartcatalytec
ball mill initial charge chart Optimization of mill performance by usingSciELO. Stresses generated in the ball charge increase which may result in spalling of balls and blocking the grate discharge. Optimization of mill performance by using online ball and pulp measurements by B. Clermont and B. de Haas Synopsis Ball mills are usually
• ### Charge For The Ball Mill- Jaw crusher ball mill Mining
Charge For The Ball Mill. The density of the charge must account for all of the material in the mill including the media which may be steel balls in a ball mill or large lumps of ore in an autogenous mill or a mixture in a semiautogenous mill as well as the slurry that makes up th Charge for the ball mill.
• ### calculating steel ball charge in ball mill
Exploring ball size distribution in coal grinding millsResearchGate. ABSTRACT Tube mills use steel balls as grinding media. new balls periodically to maintain a steady balanced ball charge in the mill. .. of the calculation of the size distribution of the equilibrium mixture of balls in a ball mill is developed.
• ### What is a ball mill What are its uses and advantages Quora
Mar 18 2018 · Ball Mill The rotation of the Ball mill causes the charge consisting of grinding media (steel balls) and feed material to be lifted due to centrifugal forces and friction between the media and the lining. The height to which the charge is lifted d
• ### ball mill initial charge calcProductsKefid Machinery
ball mill initial charge calc. 250tph river stone crushing line in Chile. 200tph granite crushing line in Cameroon. 250tph limestone crushing line in Kenya. 250tph granite crushing line in South Africa. 120tph granite crushing line in Zimbabwe. 400tph crushing plant in Guinea. Chat Online
• ### What is a ball mill What are its uses and advantages Quora
Mar 18 2018 · Ball Mill The rotation of the Ball mill causes the charge consisting of grinding media (steel balls) and feed material to be lifted due to centrifugal forces and friction between the media and the lining. The height to which the charge is lifted d
• ### Optimization of mill performance by using
Stresses generated in the ball charge increase which may result in spalling of balls and blocking the grate discharge. Optimization of mill performance by using online ball and pulp measurements by B. Clermont and B. de Haas Synopsis Ball mills are usually the largest consumers of energy within a mineral concentrator.
• ### how to charge media in ball millws-auto
Number size and mass of each ball size depends on mill load and whether or not the media is being added as the initial charge. For the initial chargin of a mill Coghill and DeVaney (1937) defined the ball size as a function of the top size of the feed i.e.
• ### Vertimill®
It is also faster to install than a traditional ball mill significantly reducing initial costs. Lower operating cost. Vertimill® is an energy efficient grinding machine. They tend to grind more efficiently than for example ball mills with feeds as coarse as 6 mm to products finer than 20 microns.
• ### Ball Mill Critical Speed Working PrincipleYouTube
Jun 20 2015 · Learn about Ball Mill Critical Speed and its effect on inner charge movements. The effect of Ball Mill RPM speed going from sub-critical to super-critical helps understand the Ball Mill Working
• ### AMIT 135 Lesson 7 Ball Mills CircuitsMining Mill
Ball Size as Initial Charge. Commercial ball sizes 10150 mm Number size and mass of each ball size depends on mill load and whether or not the media is being added as the initial charge. For the initial chargin of a mill Coghill and DeVaney (1937) defined the ball size as a function of the top size of the feed i.e. d↓V = 0.40 K√F
• ### Ball mill initial charge calcbluetowerscommunity
Ball Size as Initial Charge. Commercial ball sizes 10150 mm Number size and mass of each ball size depends on mill load and whether or not the media is being added as the initial charge. For the initial chargin of a mill Coghill and DeVaney (1937) defined the ball size as a function of the top size of the feed i.e. d↓V = 0.40 K√F
• ### Difference Between Sag Mill vs Ball Millmech4study
Oct 12 2017 · Ball mill is a fine grinder. A horizontal or vertical rotating cylinder which is filled partially with the balls of ceramics small rocks and balls made from stainless steel. The ball charge of a SAG mill is about 29 to 30 . By friction and influence of tumbling balls inside rotating cylinder grinds the raw material to the required fineness.
• ### ball mill initial charge calculationmineralenzoeken
calculate ball mill chargeeitmindiacoin calculation of ball mill charge volume - How to Calculate Charge Volume in Ball or Rod MillMining and The charge volume of a ball or rod mill is expressed as the percentage of the volume within the 24/7 online ball mill initial charge charticchmt2017org
• ### NEW LOCKED CHARGE PROTECTION SYSTEM PREVENTS
the mill charge and circulating load. This is when locked charge damage is most likely and where normal barring procedure has not been effective in eliminating locked charges. It can be seen that the Ball mill tumbling angle is typically very high with a total of 8 start-ups
• ### Optimum choice of the make-up ball sizes for maximum
A grinding circuit simulation combined with ball weal law was used to determine the optimum composition of the make-up ball sizes in tumbling ball mills. It was found that the optimum composition depends on various factors including the feed size the product size the mill diameter and the breakage parameters.
Ball Mill LoadingWet Milling. Ball Mill Loading (wet milling) When charging a ball mill ceramic lined mill pebble mill jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product.
• ### ball mill initial chargestadsschoonarnhem
ball mill initial charge led with larger diameter grinding media and lined with lifting liners. The first chamber coarse-grinds the feed material and prepares it for the second chamber. The second chamber is the fine grinding chamber. TECHNICAL NOTES 8 GRINDING R. P. King Mineral Techmaterial in the mill including the media which may be steel balls in a ball mill or large lumps of ore in
• ### Stirred Milling VERTIMILL® Grinding Mills Stirred Media
Stirred Mills have been proven to provide energy savings compared with traditional ball mills. The finer the product required the more efficient stirred mills will be than a ball mill. The attrition grinding action vertical arrangement and the finer media size distribution contribute to make stirred mills more energy efficient grinding machines.
• ### THE OPTIMAL BALL DIAMETER IN A MILL
determining the optimal ball charge in mills. In the first part of the paper on the basis of the theoretical analysis of the energy-geometric correlations which are being established during the grain comminution by ball impact as well as on the basis of the experiment carried out on
• ### Ball charges calculatorsthecementgrindingoffice
- Ball top size (bond formula) calculation of the top size grinding media (balls or cylpebs) -Modification of the Ball Charge This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency
• ### Optimization of Cement Grinding Operation in Ball Mills
Operation and Elements of a Closed Circuit Ball Mill System. Cement ball mills typically have two grinding chambers. The first chamber is filled with larger diameter grinding media and lined with lifting liners. The first chamber coarse-grinds the feed material and prepares it for the second chamber. The second chamber is the fine grinding chamber.
• ### ball mill initial charge chartcentrobembo
Ball Mill Initial Charge Chart- Jaw crusher ball mill Ball Mill Initial Charge Chart. The same policy is employed for the makeup charge where adding different ball sizes is done in order to optimize the mill performance the ball mill liners and grinding media employed were made from stainless steel in fig 4 the initial fractions of size Ball mill initial charge chart.
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# HackerEarth Monk's Encounter with Polynomial problem solution
In this HackerEarth Monk's Encounter with Polynomial problem solution Our monk, while taking a stroll in the park, stumped upon a polynomial ( A X2 + B X +C ) lying on the ground. The polynomial was dying! Being considerate, our monk tried to talk and revive the polynomial. The polynomial said:
I have served my purpose, and shall not live anymore. Please fulfill my dying wish. Find me the least non-negative integer Xo, that shall make my value atleast K i.e., A Xo2 + B Xo + C >= K .
Help our Monk fulfill the polynomial's dying wish!
## HackerEarth Monk's Encounter with Polynomial problem solution.
`#include <bits/stdc++.h>using namespace std;#define ll long long intint main(){ int test; cin>>test; while(test--) { ll A,B,C,K; cin>>A>>B>>C>>K; ll lb=0, ub =100000; ll ans = -1; while(lb<=ub) { ll mid = (lb+ub)/2; ll val = A*(mid*mid) + B*mid + C; if(val >= K){ ans = mid; ub = mid-1; } else lb = mid+1; } cout<<ans<<endl; } return 0;}`
### Second solution
`#include<bits/stdc++.h>using namespace std;long long a,b,c;long long f(long long x){ return a*x*x + b*x + c;}int main(){ int t; long long k; cin>>t; assert(1 <= t && t <= 100000); while(t--) { cin>>a>>b>>c; cin>>k; assert(1 <= a && a <= 1000000); assert(1 <= b && b <= 1000000); assert(1 <= c && c <= 1000000); assert(1 <= k && k <= (long long)1e10); int ans = 100000000; int l = 0 , h = -b/(2*a) , m; while(l <= h) { m = (l+h)/2; if(f(m) >= k) { ans = min(ans,m); l = m + 1; } else { h = m - 1; } } l = max(0LL,-b/(2*a)) , h = 10000000; while(l <= h) { m = (l+h)/2; if(f(m) >= k) { ans = min(ans,m); h = m - 1; } else { l = m + 1; } } cout<<ans<<endl; } return 0;}`
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Solution Manual for Financial Accounting 9th Edition by Harrison Horngren and Thomas Link full download: https://www.testbankfire.com/download/solution-manualfor-financial-accounting-9th-edition-by-harrison-horngren-andthomas/ Chapter 2: Transaction Analysis Learning Objective 1: Explain what a transaction is 1. What is a transaction? Listed below are some events. Indicate which of the following events would be considered a transaction of Baskin Real Estate Services, Inc. Explain your answers.
A transaction is any event that has a financial impact on the business and can be measured reliably. a. Richard Baskin inherited \$100,000 from his crazy uncle Fred. Not a transaction of the business b. Richard invested the \$100,000 inheritance in his business, Baskin Real Estate Services Inc., and received common stock in exchange. Impacts business and can be measured
c. Richard hired an office manager. The office manager will receive a salary of \$1,800 every two weeks. Impacts business, but not financially d. Richard paid the office manager, \$1,800. Impacts business and can be measured e. Richard met with a client who agreed to pay Richard \$1,000 a month for managing some rental property. Impacts business, but no money has changed hands and Richard hasn’t done anything yet. f. Richard billed the client \$1,000 for January services on January 31. Impacts business and can be measured g. Richard received a check from the client on February 15. Impacts business and can be measured Learning Objective 2: Define “account,” and list and differentiate between different types of accounts 1. What is an account? An account is the record of all the changes in a particular asset, liability, or stockholders’ equity during a period. 2. Every account contains the information on the left. Cash
Accounts Payable
Beg. Balance + Increases
Cash receipts
Purchases on credit Chapter 2 Transaction Analysis
2-1
- Decreases
Cash payments
Payments of accounts payable
= End. Balance a. What transactions cause Cash to increase? Which transactions cause Cash to decrease? Enter your answers in the table above. b. What transactions cause Accounts Payable to increase? Which transactions cause Accounts Payable to decrease? Enter your answers in the table above.
3. Choose a company whose product that you buy (such as Prentice Hall, the publisher of this textbook). List two asset accounts, two liability accounts, two stockholders’ equity accounts, two revenue accounts, and two expense accounts that your company might have. Answers will vary. Instructors can use this question to compare the similarities and differences of companies.
Learning Objective 3: Show the impact of business transactions on the accounting equation Indicate the effect of the following transactions on the accounting equation in the chart below. Transaction (a) has been provided as an example. a. Received \$500,000 from the owners as an investment in the company in exchange for common stock. (pg 63) b. Purchased \$2,000 of supplies on account. (pg 64) c. Purchased land for \$60,000, paying \$10,000 down and the balance on a note payable. d. Performed services on account, \$4,000. (pg 64) e. Paid \$5,000 for salaries.(pg 65)
Cash a.
Accounts Receivable
d.
Land
Accounts Payable
+500,000
Common Stock
Retained Earnings
+500,000
b. c.
Supplies
Notes Payable
+ 2,000 (10,000)
+ 2,000 + 60,000
+ 4,000
+50,000 + 4,000
e.
(5,000)
(5,000)
Chapter 2 Transaction Analysis
2-3
IN-CLASS EXAMPLE Analyze the effects of the transactions given below as illustrated in transaction 1. ASSETS = LIABILITIES Accounts Notes Accounts Cash Receivable Supplies Truck Payable Payable Bal
\$ 10,000
1.
210,000
\$ 2,000
\$ 500
\$ 2,500
+ 1,000 (1,500)
4.
5,000
5.
STOCKHOLDERS’ EQUITY Common Retained Stock Earnings \$ 4,000
\$
+ 1,000 + 12,000
+ 10,500 + 5,000
+ 1,000
6.
(1,200)
7.
500
8.
(700)
9.
(1,000)
10.
(800) \$20,300
+1,000 (1,200)
(500) (700) (1,000)
(1,000) (800)
\$2,500
\$1,500
1. Boring invested \$210,000 in exchange for common stock. 2. Purchased \$1,000 of supplies on account. 3. Purchased a truck for \$12,000, terms \$1,500 down and the balance on a 2-year note payable. 4. Provided \$5,000 of services for cash.
2-4
6,000
+210,000
2. 3.
+
Financial Accounting 9/e Instructor’s Manual
\$12,000
\$10,500 5. 6. 7. 8. 9.
\$2,500
\$214,000
Provided \$1,000 of services on account. Paid an employee salary of \$1,200. Collected \$500 on account Paid a \$700 dividend. Paid \$1,000 on account for the supplies purchased in transaction 2. 10. Paid \$800 rent.
\$9,300
Learning Objective 4: Analyze the impact of business transactions on accounts 1. Enter the words debit, credit, increase or decrease on the appropriate sides of the T-accounts shown below. Refer to page 73for guidance. What is the same? What is different?
Debit Increase
Assets Credit Decrease
Debit Decrease
Liabilities Credit Increase
Owners’ Equity Debit Credit Decrease Increase
Debit and credit is the same for all types of accounts; increases and decreases are on different sides.
2. Certain T-accounts are listed below. Indicate if the account is an asset, liability, or stockholders’ equity account. Enter the words debit, credit, increase or decrease on the appropriate sides of the T-accounts. Then, draw a circle around those accounts where increase are recorded on the left side of the account. An example is provided for you. Cash (A) Accounts Receivable (A) Supplies (A) Debit Increase
Credit Decrease
Debit Increase
Credit Decrease
Truck (A) Debit Credit Increase Decrease
Note payable (L) Debit Credit decrease increase
Service Revenue (SE) Debit Credit Decrease Increase
Salary Expense (SE) Debit Credit Decrease Increase
Common Stock (SE) Debit Credit Decrease Increase
Dividends (SE) Debit Credit Increase Decrease
Debit Increase
Credit Decrease
Accounts Payable (L) Debit Credit Decrease Increase
Retained Earnings (SE) Debit Credit Decrease Increase
3. Refer to question 1 under Objective 3. Indicate (1) the accounts involved in each transaction, (2) if the account is an asset (A), liability (L), stockholders’ equity (SE), revenue (R), or expense (E), (3) if the account increases or decreases, and (4) if the accounts are debited or credited. The first one is provided as an example
Account a. Cash Common Stock b. Supplies
A, L, SE, R or E
Increase/Decrease
Debit/ Credit
A
Increase
Debit
SE
Increase
Credit
A
Increase
Debit
Chapter 2 Transaction Analysis
2-5
Accounts Payable
L
Increase
Credit
c. Land
A
Increase
Debit
Cash
A
Decrease
Credit
Notes Payable
L
Increase
Credit
d. Accounts Receivable
A
Increase
Debit
Service Revenue
R
Increase
Credit
e. Salaries Expense
E
Increase
Debit
A
Decrease
Credit
Cash
Learning Objective 5: Record (journalize and post) transactions in the books 1. What information is recorded in the journal? Why is it helpful to have information recorded in this way? The journal is a chronological record of each transaction. The journal records the debit and the credit together to ensure the equality of the accounting equation as well as a memo for each entry. If an accountant needed to look up a transaction that occurred on a particular date, it would be easy to find that information in the journal. 2. Using the information from the IN-CLASS EXAMPLE in Learning Objective 4, prepare the journal entries for first two transactions. The remaining items will be reviewed in class.
IN-CLASS EXAMPLE Date
Account Name
1
Cash
Debit
Credit
210,000 Common Stock
210,000
Issued common stock.
2
Supplies
1,000
Accounts Payable
1,000
Purchased supplies on credit.
3
Truck
12,000 Cash
1,500
Notse Payable 2-6
Financial Accounting 9/e
10,500 Instructor’s Manual
Purchased truck for cash and a note. 4
Cash
5,000 Service Revenue
5,000
Provided services for cash.
5
Accounts Receivable
1,000
Service Revenue
1,000
Provided services on account.
6
Salary Expense
1,200
Cash
1,200
Paid salaries.
7
Cash
500 Accounts Receivable
500
Collected cash on account.
8
Dividends
700
Cash
700
Paid dividends.
9
Accounts Payable
1,000
Cash
1,000
Paid cash on account.
10
Rent Expense
800
Cash
800
Paid rent.
Chapter 2 Transaction Analysis
2-7
3. What information is recorded in the general ledger? Why is it helpful to have information recorded in this way? How is this information different than the information in the journal? The journal organizes information by date while the ledger organizes information by account. The ledger shows the beginning balance of an account, the increases and decreases (debits and credits), and the ending balance. The ending balance is reported on the financial statements. Without the journal, the equality of debits and credits would be very difficult to verify and without the ledger, the balances of the accounts would be difficult to compute. 4. Using the information in the journal above, post the first two transactions to the general ledger. The remaining items will be reviewed in class. IN-CLASS EXAMPLE Cash Bal. 1. 4. 7.
10,000 210,000 5,000 500
3. 6. 8. 9. 10.
Accounts Receivable 1,500 1,200 700 1,000 800
Bal. 220,300
Bal. 2,000 5. 1,000
500
Bal. 2,500
Truck 4.
7.
12,000 9.
Bal. 1,500 Notes Payable 3. 10,500
Bal. 2,500
Common Stock Bal 4,000 1.
Retained Earnings Bal. 6,000
Bal. 10,500
8.
Dividends 700
Bal.
700
210,000
Bal. 214,000
Bal. 6,000
Revenues
Salary Expense
4
5,000
5.
1,000
Bal. 6,000
Rent Expense
6.
1,200
10.
800
Bal.
1,200
Bal.
800
Learning Objective 6: Construct and use a trial balance What is a trial balance? When is it prepared? 2-8
Bal 500 2. 1,000
Accounts Payable Bal. 2,500 1,000 2. 1,000
Bal. 12,000
1.
Supplies
Financial Accounting 9/e Instructor’s Manual
A trial balance is a listing of all of the balances from the accounts. In a real accounting system, you couldn’t see all of the accounts at the same time, so the trial balance allows you to see all of the account balances in one place. A trial balance can be prepared at any time. 2.
Using the information from the T-accounts above, fill in the trial balance.
Trial Balance Boring Company 12/31/x1 Credit
Debit Cash
\$
220,300
Accounts Receivable
2,500
Supplies
1,500
Truck
12,000
Accounts Payable
\$
2,500
Notes Payable
10,500
Common Stock
214,000
Retained Earnings
6,000
Dividends
700
Revenues
6,000
Salary Expense
1,200
Rent Expense
800 \$
239,000
\$
239,000
IN-CLASS EXAMPLE A. What would the totals of the trial balance be the debit to Accounts Receivable had not been posted in transaction 5? Would Accounts Receivable be overstated, understated, correctly stated? The debit column total would be \$1,000 (understated) because the Accounts Receivable account is understated by \$1,000. The credit column totals would still be \$239,000. To find this error, compute the difference between the columns and look for an entry for that amount.
Chapter 2 Transaction Analysis
2-9
B. What would the totals of the trial balance be if transaction 5 had been posted as a debit to Cash and a credit to Service Revenue. Would Accounts Receivable, Cash, and Service Revenue be overstated, understated, correctly stated? The debit and credit column totals would still be \$239,000. Cash would be overstated \$1,000 and Accounts Receivable would be understated by \$1,000. Since both of these accounts are on the debit side of the trial balance, the trial balance total is not affected. Service Revenue is correct. C. What would the totals of the trial balance be if transaction 5 had been posted as a credit to Accounts Receivable and a credit to Service Revenue. Would Accounts Receivable, Cash, and Service Revenue be overstated, understated, correctly stated? The debit column would be understated by \$2,000 because Accounts Receivable is understated by \$2,000. The credit column total would still be \$239,000. Service Revenue is correct. To find this error, compute the difference between the column totals (\$2,000) and divide by 2. Cash is correct. D. What would the totals of the trial balance be if transaction 5 had been posted as a debit to Accounts Receivable for \$1,000 and credit to Service Revenue for \$100. Would Accounts Receivable, Cash, and Service Revenue be overstated, understated, correctly stated? The debit column and Accounts Receivable would be correctly stated. The credit column would be understated by \$900 because Service Revenue is understated by \$900. To find this error, compute the difference between the two columns and if the difference is divisible by 9, then the error might either be a slide (adding too many or too few zeros) or a transposition error (rearranging the digits in a number). Cash is correct.
3. Using the following information, compute the missing information. Prepare T-accounts to help. How does the normal balance of the account affect your solution? Cash \$ 10,000 160,000 150,000 \$ 20,000
Beginning balance Receipts Payments Ending balance
Beginning balance Revenue on account Collections on account Ending balance
Accounts Receivable
Cash
Bal. Receipts Bal.
2-10
10,000 160,000 20,000
Payments 150,000
Accounts Receivable \$ 25,000 240,000 244.000 \$ 21,000
Bal. Revenue Bal.
Financial Accounting 9/e Instructor’s Manual
25,000 240,000 Collections 244,000 21,000
The normal balance of an account is on the side on which increases are recorded. Cash and Accounts Receivable are both assets and have a normal debit balance.
Chapter 2 Transaction Analysis
2-11
# Solution manual for Financial Accounting 9th Edition by Harrison Horngren and Thomas
Link full download: https://bit.ly/2EcvtZW Language: English ISBN-10: 0132751127 ISBN-13: 978-0132751124 ISBN-13: 9780132751124 Financial A...
# Solution manual for Financial Accounting 9th Edition by Harrison Horngren and Thomas
Link full download: https://bit.ly/2EcvtZW Language: English ISBN-10: 0132751127 ISBN-13: 978-0132751124 ISBN-13: 9780132751124 Financial A...
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# What Happens When B Is Negative?
## What is the period of sin?
The period of the sine curve is the length of one cycle of the curve.
The natural period of the sine curve is 2π.
So, a coefficient of b=1 is equivalent to a period of 2π.
To get the period of the sine curve for any coefficient b, just divide 2π by the coefficient b to get the new period of the curve..
## What happens if the amplitude is negative?
The difference with a negative value of a however, is our sine curve now has a negative amplitude. In other words, our graphs are the same as when a was a positive value, but are now reflected across the x-axis. To see a direct comparison between positive and negative values of a, click here.
## What does negative b over 2a find?
To find the vertex, we can use -b over 2a. … The negative on the x² tells us our parabola is facing downwards so we have a y coordinate of our vertex at -1 and our parabola goes downwards, sorry the y coordinate of the vertex at 1, parabola facing downwards, so we know that 1 is going to be the largest y value get.
## What does a negative cosine graph look like?
Hare – sign is in front of the cosine graph. If we draw the negative cosine it will produce the reflection about the x – axis. So the negative cosine graph will be opposite to the positive graph.
## How do you self check your stomach for pregnancy?
Walk your fingers up the side of her abdomen (Figure 10.1) until you feel the top of her abdomen under the skin. It will feel like a hard ball. You can feel the top by curving your fingers gently into the abdomen. Figure 10.1 With the woman lying on her back, begin by finding the top of the uterus with your fingers.
## What’s the difference between sin and cos graphs?
In a cosine graph, a positive or negative number vertically flips the graph and determines whether the graph starts at the maximum (if it’s positive) or minimum (if it’s negative). For a sine graph, a positive or negative number vertically flips the graph like it does with a cosine graph.
## How do you know if the amplitude is negative?
Amplitudes are always positive numbers (for example: 3.5, 1, 120) and are never negative (for example: -3.5, -1, -120). Amplitudes are positive because distance can only be greater than zero or equal to zero; negative distance does not exist.
## What is the formula for period?
The formula for time is: T (period) = 1 / f (frequency). λ = c / f = wave speed c (m/s) / frequency f (Hz). The unit hertz (Hz) was once called cps = cycles per second.
## How do u calculate period?
To figure out how long your cycle is, start at cycle day 1 of your last menstrual cycle and begin counting (Cycle day 1,2,3,4 and so forth). The length= the last cycle day before you started bleeding again.
## What if the discriminant is negative?
The discriminant can be positive, zero, or negative, and this determines how many solutions there are to the given quadratic equation. A positive discriminant indicates that the quadratic has two distinct real number solutions. … A negative discriminant indicates that neither of the solutions are real numbers.
## Why is my period late if I’m not pregnant?
Pregnancy is by far the most common cause of a missed period, but there are some other medical reasons and lifestyle factors that impact your menstrual cycle. Extreme weight loss, hormonal irregularities, and menopause are among the most common causes if you’re not pregnant.
## How do you tell if a graph is sine or cosine?
Look at the graphs of the sine and cosine functions on the same coordinate axes, as shown in the following figure. The graph of the cosine is the darker curve; note how it’s shifted to the left of the sine curve. The graphs of y = sin x and y = cos x on the same axes.
## Where do sine graphs start?
origin pointThe Sine curve started at the origin point. An easy way to remember these graphs is to know their 5 important points. The zeros, maximum and minimum points. The Sine curve has zeros at the beginning, middle and end of a cycle.
## What does a negative period mean?
The period of the graph is , as is the period of . The effect of the negative sign on the inside is to replace x-values by their opposites. This will flip the graph around the y-axis. However, because the graph of cosine is symmetric about the y-axis, this has no effect at all.
## Can I still be pregnant if the test is negative and no period?
Your hCG levels will increase over time if you are pregnant. A negative pregnancy test result means the test hasn’t detected hCG in your urine. If your period is very late, or you’ve skipped your period, and you get a negative result, you are unlikely to be pregnant.
## What does B do in a quadratic equation?
Quadratic function: The quadratic function is f(x) = a * x^2 + b * x + c, which tells you what the function will look like graphed. B-value: The b-value is the middle number, which is the number next to and multiplied by the x; a change in the value of b affects the parabola and the resulting graph.
## Can a frequency be negative?
Negative frequency is an idea associated with complex exponentials. A single sine wave can be broken down into two complex exponentials (‘spinning numbers’), one with a positive exponent and one with a negative exponent. That one with the negative exponent is where you get the concept of a negative frequency.
## What if B is negative in a quadratic formula?
This relationship is always true: If you get a negative value inside the square root, then there will be no real number solution, and therefore no x-intercepts. In other words, if the the discriminant (being the expression b2 – 4ac) has a value which is negative, then you won’t have any graphable zeroes.
## What is B in a sine function?
The number, B, in front of x is number of cycles seen in 0 to 2π interval. The value B is the number of cycles the graph completes in an interval of from 0 to 2π or 360º. The value B affects the period. The period of sine and cosine is. .
## Why is my period late not pregnant?
Your cycle Missed or late periods happen for many reasons other than pregnancy. Common causes can range from hormonal imbalances to serious medical conditions. There are also two times in a woman’s life when it’s totally normal for her period to be irregular: when it first begins, and when menopause starts.
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# What is the analytical form of the cylindrical wave appearing on reflection of a plane wave from a corner?
This is a cross-post from Math.SE, where no answer was given after 3 months.
Consider a plane 2D wavelet moving towards a corner reflector with 120° opening angle with infinitely extended sides. The surface of the reflector has homogeneous Dirichlet boundary conditions imposed, and the wave obeys the usual hyperbolic wave equation:
$$\partial_{tt} u(x,y,t)=\partial_{xx}u(x,y,t)+\partial_{yy}u(x,y,t).$$
The solution, with the initially propagating part and the reflections, can be easily constructed by rotating the initial wavelet, changing its sign and putting the resulting wavelet next to the initial one so as to satisfy the boundary conditions by canceling the wave function at the boundaries. The result will look like this (ignore the small-wavelength artifacts, they are due to numeric errors in the simulation):
But as the points of the slanted reflected waves come close to the corner, there appears a problem: simply "sliding" the reflected part no longer works, since the shape of the reflecting boundary abruptly changes. Moreover, numerical simulation (see below) shows that the reflection from the corner doesn't produce a backwards-propagating plane wave (that we'd get from a 90° corner): instead the original slanted reflections continue their paths, and a new, cylindrical, wave originates from the corner. This cylindrical wave appears to cancel out the values of the slanted reflections on the boundary to satisfy the boundary conditions.
My question is: what is the analytical form of this cylindrical wave? It doesn't seem to be a Bessel function, because Bessel functions don't have constant amplitude nor constant wavelength (they change with radius). So what is it then? Does it have a closed form? Or is it at least explicitly expressible as an integral or a series?
The keywords to search for this "scattering by a corner reflector" appear to have been "diffraction by a wedge". There's been quite a bit of research on exactly this kind of boundary. Particularly, ref. 1 provides explicit expressions for the solution of the problem that's set up as follows.
A two-dimensional wedge is situated in the space $$0\le r<\infty$$, $$\alpha\le \theta\le 2\pi$$, with $$(r,\theta)$$ being cylindrical coordinates (omitting $$z$$), as shown in the figure below (figure 1 from ref. 1). A plane wave is incident from the direction $$\theta=\theta_0$$ and scattered by the inner part of the wedge, which may be of any opening angle from $$0$$ to at least $$4\pi$$ (the case of hard-soft semi-infinite plane, see ref. 1 p. 277).
The solution of the wave equation, i.e. incident+scattered field, corresponding to $$\alpha=120°$$ is given in $$\S 4$$ as $$G_{2\pi/q}(r,\theta,\theta_0;k)$$ by equation $$(23)$$; in our case $$q=3$$. The expression (tweaked here for presentation) is
$$\begin{multline} G_{2\pi/q}(r,\theta,\theta_0;k)=\sum_{m=0}^{q-1}\sum_{N\in\mathbb Z} \operatorname{H}[\pi-|\Delta\theta_m+4\pi N|]e^{ikr\cos(\Delta\theta_m)}+\\ +\frac{e^{i\pi/4}}{\sqrt\pi}\sum_{m=0}^{q-1} e^{ikr\cos(\Delta\theta_m)} \operatorname{sgn}\left[ \cos\left(\Delta\theta_m/2\right)\right] \intop_{+\infty}^{\left| \cos\left(\Delta\theta_m/2\right) \right| \sqrt{2kr}} e^{-iv^2}\,\mathrm{d}v, \end{multline}$$
where $$\Delta\theta_m=\theta-\theta_0+4\pi m/q,$$ $$\operatorname{H}[\cdot]$$ is the Heaviside step function, and $$\operatorname{sgn}(\cdot)$$ is the signum function. The summation over $$N$$ will pick only a handful of terms, because for others the Heaviside function will vanish. The integral in the second sum is expressible in terms of the Fresnel sine and cosine integrals:
$$\int\limits_{+\infty}^x \exp(-iv^2)\,\mathrm{d}v= \frac{1-i}2 \sqrt{\frac\pi2} \left( (1+i)\operatorname{C}\left(\sqrt{\frac2\pi}x\right) + (1-i)\operatorname{S}\left(\sqrt{\frac2\pi}x\right)-1\right).$$
The cylindrical wave asked about in the OP originates from the second sum with these Fresnel-integral terms.
### References
1. A. D. Rawlins, Plane-wave diffraction by a rational wedge, Proc. R. Soc. A 411, 265-283 (1987).
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Search All of the Math Forum:
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Topic: Help Needed in math
Replies: 25 Last Post: May 2, 2014 6:01 PM
Messages: [ Previous | Next ]
Karl-Olav Nyberg Posts: 1,417 Registered: 12/6/04
Re: Help Needed in math
Posted: Apr 29, 2014 6:06 PM
On Monday, April 28, 2014 10:30:47 AM UTC+2, punisher wrote:
> --------------------------------------------------------------------------------------------------------
> YOU CAN REMOVE THIS TEXT MESSAGE BY BEING A PAID MEMBER FOR \$10/year.
> Message ID= 113107
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>
>
>
> Can someone help me solve this
>
> Find the largest values of the integer a for which ax^2-5x-3 is negative for all values of x.
>
>
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> YOU CAN REMOVE THIS TEXT MESSAGE BY BEING A PAID MEMBER FOR \$10/year.
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Hi.
set a = (5x + 3) / x^2 and find the minima of this function. This will, with your constraints, give the answer: a = -3
KON
Date Subject Author
4/28/14 punisher
4/28/14 quasi
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/28/14 Peter Percival
4/29/14 William Elliot
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
5/1/14 Karl-Olav Nyberg
5/1/14 William Elliot
5/1/14 quasi
5/2/14 William Elliot
5/2/14 quasi
5/2/14 William Elliot
5/2/14 snmpprotocol@gmail.com
5/2/14 quasi
5/2/14 Karl-Olav Nyberg
5/2/14 quasi
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✊🏿 Black Lives Matter. Please consider donating to Black Girls Code today.
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# Plotly Scattermapbox positive to negative longitudes
Hello! I am trying to make a scattermapbox using plotly and am running into some issues:
I am trying to plot a filled area using the following coordinates:
lats = [17,7, 13, 15, 17]
lons = [-170, 100, 136, 153, -170]
The problem I am running into is how go.Scattermapbox is displaying the lines. Instead of drawing a line connecting longitude 100 to -170 and crossing the 180 meridian, it draws the line all the way around the world (Please view the attached image). I believe this is due to my points straddling -180 degrees longitude. If I just add 360 to the 170, it is displayed correctly, however that just moves the problem to values that straddle 0 degrees longitude (prime meridian). Has anyone else encountered this? How do you fix this issue?
Hi @dm1681,
Reorder your data, and replace -170 by 190 (if we are identifying 180 with -180, then -170 gets 190).
lat = [7, 13,15,17 ]
lon= [100, 136, 153, 190]
With these data you’ll get the right plot:
``````fig=go.Figure(go.Scattermapbox(lat=lat, lon=lon, fill='toself'))#190 represents -170
fig.update_layout(height=600,
mapbox=dict(
accesstoken='',
bearing=0,
center=dict(
lat=12,
lon=145 ),
pitch=0,
zoom=1,
style='basic'
)
)
fig.show()
``````
Eventually you can add customdata and hovertemplate to get displayed on hover the right lon ,-170, instead of 190.
.
Yes, I had already come to that solution (by adding 360) and mentioned it in my post. I was not sure how to apply that solution generically. For instance, if I do it for every longitude < 0, then the same problem persists, except straddling 0 degrees longitude.
After some searching, I found that in order to determine which points needed to be mapped, I calculate relative distances between all the points and find the point that has the average distance > 180. Then if that point is negative, I add 360, if it is positive, I subtract 360.
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문제1074--Longest Ordered Subsequence
### 1074: Longest Ordered Subsequence
실행시간 제한: 1 Sec 메모리사용 제한: 128 MB
제출: 548 통과: 205
[제출] [채점기록] [묻고답하기]
#### 문제 설명
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
#### 입력 설명
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000
#### 출력 설명
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
#### 입력 예시 Copy
7
1 7 3 5 9 4 8
#### 출력 예시 Copy
4
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# R/CM.calculateCenterline.r In AntoniusGolly/cmgo: Derive principle Channel metrics from bank points
#### Documented in CM.calculateCenterline
#' Calculate channel centerline
#'
#' Calculate the centerline of the channel polygon in 5 steps:
#' \enumerate{
#' \item creating Voronoi polygons of the bank points, convert to paths (line segments with two ends) and remove duplicates\cr
#' \item filtering for path segments that lie within the banks\cr
#' \item filtering for path segments that are dead ends (have less than 2 connected ends)\cr
#' \item sorting of the centerline segments to generate centerline\cr
#' \item smooth the centerline
#' }
#'
#' \code{CM.calculateCenterline()} calculates the centerline of the channel polygon (Fig. 7).
#'
#' \if{html}{\figure{06-processing.png}{options: width="800px" alt="Figure: processing"}}
#' \if{latex}{\figure{06-processing.pdf}{options: width=9cm}}
#' \emph{Figure 6: A visualization of the calculation of the centerline a) the channel polygon, b) the Voronoi polygons,
#' c) extraction of the centerline segments, d) smoothing of the centerline path.}
#'
#' The function requires as input the channel polygon (Fig. 6a) which must be stored within the global data object
#' The algorithm then creates Voronoi polygons around the bank points (Fig. 6b).
#' Voronoi polygons around points denote the areas within which all points are closest to that point. The polygons
#' are disassembled to single line segments. In Fig. 6b you can already notice a centerline evolving from the segments in
#' the middle of the channel polygon. To get only these segments a filtering (Fig. 7) is applied to the Voronoi segments.
#'
#' \if{html}{\figure{07-filtering.png}{options: width="600px" alt="Figure: processing"}}
#' \if{latex}{\figure{07-filtering.pdf}{options: width=6cm}}
#' \emph{Figure 7: the filtering of the Voronoi segments: a) in blue all Voronoi segments, b) in red all segments fully within
#' the channel polygon, c) in green all segments without dead ends.}
#'
#' To retrieve only the segments that represent the centerline all segments that do not lie entirely
#' within the channel banks are removed (Fig. 7b). In a second step dead ends are removed (Fig. 7c). Dead ends are
#' segments that branch from the centerline but are not part of it.
#'
#' These centerline segments will be
#' chained to one consistent line and get smoothed (Fig. 7d). The degree of smoothing can be adjusted
#' through the parameter \code{centerline.smoothing.width} (defaults to the same value as
#' \code{bank.interpolate.max.dist}). This centerline represents the reference of the river, for which
#' length, local width and slope are calculated next. Note, that the length of the centerline has decreased
#' by the smoothing in d). It is important to understand, that the length of a river is not a well-defined measure.
#' The length of a river depends on the resolution of the bank points. Similar to
#' \href{https://en.wikipedia.org/wiki/Coast#Coastline_problem}{the coast line paradox}, the length depends on the
#' scale of the observations. Technically, a bended river is a fractal, which
#' means theoretically, the length diverges to infinity at an infinitely high resolution of the bank points.
#' However, practically there is an appropriate choice of a minimum feature size. Every user has to determine this
#' scale individually and should be aware of this choice. The decrease in length due to smoothing
#' is saved as value in the global data object under \code{cmgo.obj$data[[set]]$cl$length.factor}. A value of 0.95 means #' that the length of the smoothed centerline is 95\% the length of the original centerline paths. #' #' @template param_global_data_object #' @template param_set #' @return returns the global data object extended by the centerline data \code{$cl} for the respective data set(s)
#' @author Antonius Golly
#' @examples
#' # get demo data
#' # (find instructions on how to use own data in the documentation of CM.ini())
#' cmgo.obj = CM.ini("demo")
#'
#' # generate the polygon from the bank points
#' cmgo.obj = CM.generatePolygon(cmgo.obj)
#'
#' # calculate the centerline from the polygon
#' cmgo.obj = CM.calculateCenterline(cmgo.obj)
#'
#' # check results
#' plot.par = CM.plotPlanView(cmgo.obj)
#'
#' # change degree of smoothing, re-calculate centerline and plot
#' cmgo.obj$par$centerline.smoothing.width = 12
#' cmgo.obj = CM.calculateCenterline(cmgo.obj)
#' plot.par = CM.plotPlanView(cmgo.obj)
#'
#' @export CM.calculateCenterline
CM.calculateCenterline <- function(cmgo.obj, set=NULL){
par = cmgo.obj$par data = cmgo.obj$data
sets = if(is.null(set)) names(data) else set
notice = function(x,prim=FALSE){cat(paste((if(prim) "\n--> " else " "), x, sep=""), sep="\n")}
error = function(x){stop(x, call.=FALSE)}
alert = function(x, y=""){if(y!=""){message(paste("--> ",x, y, sep=""))}else{message(paste("--> ", x, sep=""))}}
warn = function(x){warning(x, call.=FALSE)}
plot.file = function(par){if(!par$plot.to.file) return(NULL); file.no = 0 + par$plot.index; file.name = paste(par$plot.directory, str_pad(file.no, 3, pad="0"), "_", par$plot.filename, sep=""); while(file.exists(paste(file.name, ".png", sep="")) || file.exists(paste(file.name, ".pdf", sep=""))){ file.no = file.no + 1; file.name = paste(par$plot.directory, str_pad(file.no, 3, pad="0"), "_", par$plot.filename, sep="") }; dev.copy(png, filename=paste(file.name, ".png", sep=""), width=800, height=600); dev.off(); dev.copy2pdf(file=paste(file.name, ".pdf", sep=""));}
notice("calculate centerline", TRUE)
for(set in sets){
if(is.null(data[[set]]$cl$smoothed) || par$force.calc.cl){ # notice notice(paste("calculate centerline of", set, "now..."), TRUE) reasons = c() if(is.null(data[[set]]$cl$smoothed)) reasons = append(reasons, "data does not exis") if(par$force.calc.cl) reasons = append(reasons, "calculation is forced")
notice(paste("reason for calculation:", paste(reasons, collapse = " + ")))
# check if parameter matches data base
if(isTRUE(data[[set]]$polygon.bank.interpolate.max.dist != par$bank.interpolate.max.dist)){
notice(paste("interpolate max. dist of data:", data[[set]]$polygon.bank.interpolate.max.dist)) notice(paste("interpolate max. dist of par: ", par$bank.interpolate.max.dist))
warn("there is a mismatch between the current parameter par$bank.interpolate.max.dist and the dense polygon: re-run CM.generatePolygon()."); return(list( data = data, par = par)); } # check if parameter matches data base if(isTRUE(data[[set]]$polygon.bank.reduce.min.dist != par$bank.reduce.min.dist)){ notice(paste("reduce min. dist of data:", data[[set]]$polygon.bank.reduce.min.dist))
notice(paste("reduce min. dist of par: ", par$bank.reduce.min.dist)) warn("there is a mismatch between the current parameter par$bank.reduce.min.dist and the dense polygon: re-run CM.generatePolygon()."); return(list( data = data, par = par));
}
### find start/end of centerline (interpolated bank points)
cl.start = list(
x = (head(data[[set]]$channel$x[data[[set]]$ix.l], n=1) + head(data[[set]]$channel$x[data[[set]]$ix.r], n=1)) / 2,
y = (head(data[[set]]$channel$y[data[[set]]$ix.l], n=1) + head(data[[set]]$channel$y[data[[set]]$ix.r], n=1)) / 2
)
cl.end = list(
x = (tail(data[[set]]$channel$x[data[[set]]$ix.l], n=1) + tail(data[[set]]$channel$x[data[[set]]$ix.r], n=1)) / 2,
y = (tail(data[[set]]$channel$y[data[[set]]$ix.l], n=1) + tail(data[[set]]$channel$y[data[[set]]$ix.r], n=1)) / 2
)
# create Voronoi/Dirichlet/Thiessen polygons
notice("step 1 of 5: get voronoi polygons...", TRUE)
if(is.null(data[[set]]$cl$paths) || par$force.calc.voronoi || isTRUE(data[[set]]$cl$bank.interpolate.max.dist != data[[set]]$polygon.bank.interpolate.max.dist)
|| isTRUE(data[[set]]$cl$bank.reduce.min.dist != data[[set]]$polygon.bank.reduce.min.dist) ){ # notice to console notice(paste("calculate based on dense polygon with a maximum bank point distance of ", par$bank.interpolate.max.dist, sep=""))
notice(paste("calculate based on dense polygon with a minimum bank point distance of ", par$bank.reduce.min.dist, sep="")) reasons = c() if(is.null(data[[set]]$cl$paths)) reasons = append(reasons, "data does not exist") if(par$force.calc.voronoi) reasons = append(reasons, "calculation is forced")
if(isTRUE(data[[set]]$cl$bank.interpolate.max.dist != par$bank.interpolate.max.dist)) reasons = append(reasons, "max. dist. has changed") if(isTRUE(data[[set]]$cl$bank.reduce.min.dist != par$bank.reduce.min.dist)) reasons = append(reasons, "min. dist. has changed")
notice(paste("reason for calculation:", paste(reasons, collapse = " + ")))
# set meta data: bank.interpolate.max.dist
data[[set]]$cl$bank.interpolate.max.dist = par$bank.interpolate.max.dist data[[set]]$cl$bank.reduce.min.dist = par$bank.reduce.min.dist
# reset data
data[[set]]$cl$cl.paths = NULL
# calculate voronoi polygons => create paths from polygons => remove duplicated paths
voronoi = dirichlet(data[[set]]$polygon) # create voronoi polygons tiles = sapply(voronoi$tiles, "[[", "bdry") # get individual tile coordinates
paths = do.call(rbind, lapply(tiles, function(tile){ # disassemble paths of tiles
return(data.frame(
x1 = tile$x, y1 = tile$y,
x2 = c(tile$x[2:length(tile$x)], tile$x[1]), y2 = c(tile$y[2:length(tile$y)], tile$y[1])
))
}))
d.ixs = duplicated(t(apply(paths, 1, sort))) # get duplicates
data[[set]]$cl$paths = if(any(d.ixs)) paths[-which(d.ixs),] else paths # remove duplicates
rownames(data[[set]]$cl$paths) = NULL
} else {
notice("(paths taken from cache)")
}
# filter #1 (restrict to segments within banks)
notice("step 2 of 5: filter #1 (clip)...", TRUE)
if(is.null(data[[set]]$cl$cl.paths)){
in.polygon.ixs = apply(data[[set]]$cl$paths, 1, function(path){
all(point.in.polygon(path[c("x1", "x2")], path[c("y1", "y2")], data[[set]]$polygon$x, data[[set]]$polygon$y))
})
paths.in.polygon = data[[set]]$cl$paths[in.polygon.ixs,]
# from segments create points (creates duplicates)
points.in.polygon = rbind(as.matrix(paths.in.polygon[,c("x1","y1")]), as.matrix(paths.in.polygon[,c("x2","y2")]))
rownames(points.in.polygon) = NULL
colnames(points.in.polygon) = c("x", "y")
### find start/end segments from filter #1
cl.start.i = which.min((
(( cl.start$x - points.in.polygon[,"x"])^2) + (( cl.start$y - points.in.polygon[,"y"])^2)
) ^(1/2))
cl.end.i = which.min((
(( cl.end$x - points.in.polygon[,"x"])^2) + (( cl.end$y - points.in.polygon[,"y"])^2)
) ^(1/2))
# save all paths in polygon
data[[set]]$cl$paths.in.polygon = paths.in.polygon;
# save coordinates of centerline ends (start and end points of centerline)
data[[set]]$cl$ends.x = points.in.polygon[c(cl.start.i, cl.end.i), "x"]
data[[set]]$cl$ends.y = points.in.polygon[c(cl.start.i, cl.end.i), "y"]
} else {
notice("(paths.in.polygon taken from cache)")
paths.in.polygon = data[[set]]$cl$paths.in.polygon
}
# filter #2: filter out segments that have less than two connections
notice("step 3 of 5: filter #2 (dead ends)...", TRUE)
if(is.null(data[[set]]$cl$cl.paths)){
# notice
notice(paste("start iterations with", par$bank.filter2.max.it, "iterations maximum")) cl.paths = paths.in.polygon remove.continue = TRUE remove.iteration = 0 remove.ixs.first.it = NULL while(remove.continue){ # display interation and check for max remove.iteration = remove.iteration + 1 if(remove.iteration > par$bank.filter2.max.it){ warn(paste("\n### exit due to maximum iterations (max. iterations = ", par$bank.filter2.max.it, ") ###", "\nNote: this may be caused by gaps that opened in the centerline due to\njagged centerline paths. First, check for gaps visually with CM.plotPlanView(cmgo.obj, set=\"",set,"\", error=1). \nYou can than either repair these gaps by editing the centerline paths manually or \nsimply increase the bank resolution via parameter par$bank.interpolation.max.dist!", sep=""));
data[[set]]$cl$errors.filter2 = cl.paths[remove.ixs, c("x1", "y1")];
data[[set]]$cl$errors.filter2.first = cl.paths[remove.ixs.first.it, c("x1", "y1")];
data[[set]]$cl$cl.paths.tmp = cl.paths
data[[set]]$cl$errors.filter2.ixs = remove.ixs
return(list(
data = data,
par = par
))
}
cat(paste(" > iteration ", remove.iteration, ":", sep=""))
### merge all remaining cl.paths and the end points to cl.points
cl.points = rbind(
data.frame(x = c(cl.paths[,"x1"], cl.paths[,"x2"]), y = c(cl.paths[,"y1"], cl.paths[,"y2"])),
data.frame(x = data[[set]]$cl$ends.x, y = data[[set]]$cl$ends.y)
)
### calculate number of connections of each path (each and must have two)
remove.ixs = which(!(duplicated(cl.points) | duplicated(cl.points, fromLast = TRUE)))
remove.ixs[which(remove.ixs > nrow(cl.paths))] = remove.ixs[which(remove.ixs > nrow(cl.paths))] - nrow(cl.paths)
#remove.ixs = which(apply(cl.paths, 1, function(path){
# con1 = nrow(merge(cl.points, path[c("x1", "y1")])) < 2 # time intensive
# con2 = nrow(merge(cl.points, path[c("x2", "y2")])) < 2 # time intensive
# return(any(con1 + con2))
#}))
# get points to remove
if(is.null(remove.ixs.first.it)) remove.ixs.first.it = remove.ixs
if(length(remove.ixs)){
cl.paths = cl.paths[-remove.ixs, ]
notice(paste(length(remove.ixs), "elements removed!"))
} else {
notice("0 elements removed, filtering ended correctly!")
remove.continue = FALSE
}
} # while(remove.continue)
data[[set]]$cl$cl.paths = cl.paths
} else {
notice("(cl.paths taken from cache)")
cl.paths = data[[set]]$cl$cl.paths
}
# sorting
notice("step 4 of 5: sort points...", TRUE)
if(is.null(data[[set]]$cl$original)){
# cl is the final product, add two points first
cl = data.frame(x = data[[set]]$cl$ends.x[1], y = data[[set]]$cl$ends.y[1])
# ordered list of cl.paths
cl.paths.pairs = data.frame(x = c(cl.paths$x1, cl.paths$x2), y = c(cl.paths$y1, cl.paths$y2))
sort.continue = TRUE
perc = nrow(cl.paths.pairs)
perc.lev = 0
## plot percentage if more than 1000 points on centerline exists (if less it doesn't make sense since it is fast enough)
if(perc > 1000) notice("0%")
while(sort.continue){
this = as.numeric(tail(cl, n=1))
# create progress notice
perc.ac = round((perc - nrow(cl.paths.pairs)) / perc * 100)
if((perc > 1000) && (perc.ac > (perc.lev + 10))){perc.lev = perc.lev + 10; notice(paste(perc.lev,"%", sep=""))}
# find match (first point of segment)
match = which(cl.paths.pairs$x == this[1] & cl.paths.pairs$y == this[2])
# check number of machthes (must be one during sort or 0 at end)
if(length(match) == 1){
# find opposite (second point of segment)
other = if(match <= nrow(cl.paths.pairs) / 2) match + nrow(cl.paths.pairs) / 2 else match - nrow(cl.paths.pairs) / 2
# store point
cl = rbind(cl, cl.paths.pairs[other,])
# remove from paths
cl.paths.pairs = cl.paths.pairs[-c(match, other),]
} else {
if(all(this == c(data[[set]]$cl$ends.x[2], data[[set]]$cl$ends.y[2]))){
if(perc.lev < 100) notice("100%")
notice("sorting done successfully!")
# set while condition false and go to next iteration which effectively ends the while loop
sort.continue = FALSE; next
} else {
warn(paste("number of connections should be 1 but is", length(match)))
warn(paste("call CM.plotPlanView(cmgo.obj, set=\"",set,"\", error=1, error.type=\"errors.sort\") to resolve", sep=""))
data[[set]]$cl$errors.sort = matrix(this, ncol=2)
return(list(
data = data,
par = par
))
}
}
} # while(sort.continue)
# add start/end point from banks, NOTE: this is not the first/last cl path point but an interpolated point to create nicer ends
cl = rbind(cl.start, cl, cl.end)
# store
data[[set]]$cl$original = cl
} else {
notice("(cl points taken from cache)")
cl = data[[set]]$cl$original
}
cl = list(original = cl)
###########################################################
### smooth centerline #####################################
notice("step 5 of 5: smooth...", TRUE)
if(par$centerline.smoothing.width %% 2 != 1){ par$centerline.smoothing.width = par$centerline.smoothing.width + 1; notice("par$centerline.smoothing.width must be odd thus has been increased by 1")}
notice(paste("smoothing width:", par$centerline.smoothing.width)) cl$smoothed = as.data.frame(rollapply(as.data.frame(cl$original), par$centerline.smoothing.width, mean, partial=TRUE))
# add start/end point (has been moved by smoothing)
cl$smoothed[1,] = cl.start cl$smoothed[length(cl$smoothed$x),] = cl.end
###########################################################
### calculate length and cumulative length ################
notice("measure length of centerline...", TRUE)
# calculate 2d length of original centerline
diffs = diff(as.matrix(cl$original)) cl$original$seg_dist_2d = c(0, sqrt(diffs[,"x"]^2 + diffs[,"y"]^2)) cl$original$cum_dist_2d = cumsum(cl$original$seg_dist_2d) # calculate 2d length of smoothed centerline diffs = diff(as.matrix(cl$smoothed))
cl$smoothed$seg_dist_2d = c(0, sqrt(diffs[,"x"]^2 + diffs[,"y"]^2))
cl$smoothed$cum_dist_2d = cumsum(cl$smoothed$seg_dist_2d)
###########################################################
### project elevation #####################################
notice("project elevation...", TRUE)
if(!is.null(data[[set]]$channel$z)){
# take elevation from bank information if no long profile is provided (standard case)
if(is.null(data[[set]]$features$lp$x)){ notice("elevation information found: project elevation of banks to centerline", TRUE) ## on original centerline lp_closest = apply(cbind(cl$original$x, cl$original$y), 1, function(x){ return (which.min(( ((data[[set]]$channel$x - x[1])^2) + ((data[[set]]$channel$y - x[2])^2) ) ^(1/2) ) ) }) cl$original$z = data[[set]]$channel$z[lp_closest] ## on smoothed centerline lp_closest = apply(cbind(cl$smoothed$x, cl$smoothed$y), 1, function(x){ return (which.min(( ((data[[set]]$channel$x - x[1])^2) + ((data[[set]]$channel$y - x[2])^2) ) ^(1/2) ) ) }) cl$smoothed$z = data[[set]]$channel$z[lp_closest] } else { notice("elevation information found: project elevation of long profile to centerline", TRUE) ## on original centerline lp_closest = apply(cbind(cl$original$x, cl$original$y), 1, function(x){ return (which.min(( ((data[[set]]$features$lp$x - x[1])^2) + ((data[[set]]$features$lp$y - x[2])^2) ) ^(1/2) ) ) }) cl$original$z = data[[set]]$features$lp$z[lp_closest]
## on smoothed centerline
lp_closest = apply(cbind(cl$smoothed$x, cl$smoothed$y), 1, function(x){
return (which.min(( ((data[[set]]$features$lp$x - x[1])^2) + ((data[[set]]$features$lp$y - x[2])^2) ) ^(1/2) ) )
})
cl$smoothed$z = data[[set]]$features$lp$z[lp_closest] } ### calculate slope ##################################### for(local_slope_range in par$centerline.local.slope.range){
cl$smoothed[[paste("slope_fit_", local_slope_range, sep="")]] = apply(cl$smoothed, 1, function(x){
ind = which(cl$smoothed$cum_dist_2d >= x[["cum_dist_2d"]] & cl$smoothed$cum_dist_2d < (x[["cum_dist_2d"]] + local_slope_range))
fit = lm(cl$smoothed$z[ind] ~ cl$smoothed$cum_dist_2d[ind])
return(fit$coefficients[[2]]) }) cl$smoothed[[paste("slope_avg_", local_slope_range, sep="")]] = apply(cl$smoothed, 1, function(x){ ind = which(cl$smoothed$cum_dist_2d >= x[["cum_dist_2d"]] & cl$smoothed$cum_dist_2d < (x[["cum_dist_2d"]] + local_slope_range)) sl = ( tail(cl$smoothed$z[ind], n=1) - head(cl$smoothed$z[ind], n=1) ) / ( head(cl$smoothed$cum_dist_2d[ind], n=1) - tail(cl$smoothed$cum_dist_2d[ind], n=1) ) return(sl) }) } ### calculate 3d length of smoothed centerline ########## diffs = diff(as.matrix(cl$smoothed))
cl$smoothed$seg_dist_3d = c(0, sqrt(diffs[,"x"]^2 + diffs[,"y"]^2 + diffs[,"z"]^2))
cl$smoothed$cum_dist_3d = cumsum(cl$smoothed$seg_dist_3d)
} else { notice("no elevation information provided in input data: skip elevation projection") }
###########################################################
notice(paste("centerline for", set, "calculated!"), TRUE)
# store
data[[set]]$cl$original = cl$original data[[set]]$cl$smoothed = cl$smoothed
data[[set]]$cl$length.factor = tail(cl$smoothed$cum_dist_2d, n=1) / tail(cl$original$cum_dist_2d, n=1)
} else {
notice(paste("(centerline of", set, "loaded from cache)"))
}
} # for(set in names(data))
notice("CM.calculateCenterline() has ended successfully!", TRUE)
# return
return(list(
data = data,
par = par
))
}
AntoniusGolly/cmgo documentation built on Sept. 24, 2021, 1:33 a.m.
| 5,827
| 20,365
|
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| 2.90625
| 3
|
CC-MAIN-2023-06
|
latest
|
en
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https://www.hackmath.net/en/math-problem/1788
| 1,618,240,144,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00325.warc.gz
| 801,502,338
| 11,845
|
Math classification
In 3A class are 27 students. One-third got a B in math, and the rest got an A. How many students received a B in math?
n = 9
Step-by-step explanation:
$n=27\mathrm{/}3=9$
We will be pleased if You send us any improvements to this math problem. Thank you!
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| 904
| 3,562
|
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| 3.921875
| 4
|
CC-MAIN-2021-17
|
longest
|
en
| 0.96942
|
https://fr.mathworks.com/matlabcentral/cody/problems/74-balanced-number/solutions/823666
| 1,597,372,602,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00595.warc.gz
| 303,980,319
| 15,774
|
Cody
# Problem 74. Balanced number
Solution 823666
Submitted on 10 Feb 2016 by Fabricio Trentini
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 13722; assert(isequal(isBalanced(n),true))
2 Pass
%% n = 23567414; assert(isequal(isBalanced(n),true))
3 Pass
%% n = 20567410; assert(isequal(isBalanced(n),false))
4 Pass
%% n = 1; assert(isequal(isBalanced(n),true))
5 Pass
%% n = 11111111; assert(isequal(isBalanced(n),true))
6 Pass
%% n = 12345678; assert(isequal(isBalanced(n),false))
7 Pass
%% n = 12333; assert(isequal(isBalanced(n),false))
8 Pass
%% n = 9898; assert(isequal(isBalanced(n),true))
9 Pass
%% n = 469200; assert(isequal(isBalanced(n),false))
10 Pass
%% n = 57666; assert(isequal(isBalanced(n),true))
| 288
| 870
|
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| 2.890625
| 3
|
CC-MAIN-2020-34
|
latest
|
en
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https://www.robolab.in/youngs-modulus/
| 1,709,228,076,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00590.warc.gz
| 947,283,383
| 15,877
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## Young’s Modulus
Young’s modulus, also known as the elastic modulus, is a measure of the stiffness of a solid material. It is a mechanical property of linear elastic solid materials. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in a material. Young’s modulus is named after the 19th-century British scientist Thomas Young. A solid material will deform when a load is applied to it. If it returns to its original shape after the load is removed, this is called elastic deformation. In the range where the ratio between load and deformation remains constant, the stress-strain curve is linear. Not many materials are linear and elastic beyond a small amount of deformation. A stiff material needs more force to deform compared to a soft material, and an infinite force would be needed to deform a perfectly rigid material, implying that it would have an infinite Young’s modulus. Although such a material cannot exist, a material with a very high Young’s modulus can be approximated as rigid.
Thins video explains how to determine the Young’s modulus of elasticity of the material of a given wire using Searle’s apparatus.
| 236
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| 2.515625
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