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Let's consider the triangle ABC, where the distances AB and AC are constant. A and C and immobile; B can move.
The triangle can have two states:
1) There is a right angle between AB and BC, B has the same x coordinate as A
A
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|
|
|
B --- C
2) B has the same y coordinate as A, and the distance BC is double what is was before
B-----A
\
\
\
\
C
How can I calculate the position of C, in respect to A if I know the distances AB and BC?
1) known |BC| = t, |A.X - C.X| = t
2) known |AB| = f, |BC| = 2*t
unknown g = |A.Y - C.Y|
(f+t)^2 + g^2 = (2*t)^2 //grandpa Pythagoras' theorem
g = Sqrt(4*t^2 - (f+t)^2)
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I am interested in knowing how to calculate a ranking score from ratings of a product. E.g., take the apple appstore. There are two products A and B. Both have same average rating but 100 reviewers have rated A whereas 1000 reviewers have rated B. Intuitively it seems B should be ranked higher than A (it has lower standard error in the mean). Is there an established formula to compare two items and determine which is better based on their ratings?
I write some python code so you can run it easily.
def score(nn):
""" nn = [0, n1, n2, n3, n4, n5] """
if len(nn)==5:
nn = [0, *nn] # add 0
N = sum(nn)
K = 5
kk = [1,2,3,4,5]
z = 1.65 # alpha = 0.1 mean 95% confidence
avg = sum(k*(nn[k]+1)/(N+K) for k in kk)
diff = sum(k**2 * (nn[k]+1)/(N+K) for k in kk) \
- sum(k * (nn[k]+1)/(N+K) for k in kk) ** 2
dev = z * sqrt(diff / (N+K+1))
return avg - dev
You can just call it
score([5, 5, 14, 48, 223]) # 4.517059350728805
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Please find the condition below:
Col A Col B Col C Col D
235 A B Formula result
235 B
235 C
236 X
236 Y
236 Z
237 P
240 Q
I need a formula in col D to check that if Col A is 235 then Col C should have any value from A,B or C mentioned in the list in col B. if Col A is 236 then col D should give me an error if values are not between X,Y or Z.
Thanks
Enter this formula in D1: =IF(A1=235,IF(C1=B1,"Okay!","Error"),IF(OR(C1="X",C1="Y",C1="Z"),"Okay!","Error"))
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Let B be the language {0n1n | n >= 0} i.e. 0 and 1 has to have the same length
Let s in B be the string 0p1p
Assume B is regular so s must be divisible to s = xyz where xyiz i>=0 is still in B (Condition 1 of three conditions of pumping lemma).
Consider the case xyiz where i = 2 so xyyz:
Pump y with all 0s
xyyz has more 0s and 1s so it cannot be in B. Therefore, B is not regular.
I am having a hard time understanding that if y is all 0s in xyyz, then # of 0s > # of 1s
Why can't |xyy| = |z| which then it would have the same # of 0s and 1s?
'Pump y with all 0s' isn't terribly clear, but an example of how this works out is as follows:
Pick some value for y: let's say y = '0'.
Pick some value for i: i = 1
Then s = xyz. We will assume this holds true for the moment.
Now, since we assume B is regular, we know that - for any value of i - the string formed by xyiz should also be in B! Let's try xyyz, like you suggest.
...Uh-oh. You see the problem? We have to hold y constant, but doing so means we just made a string that has one more 0 than s, but doesn't have an extra 1 to go with it! We just showed that y can't be 0. Well, darn.
Now consider: is there any value of y for which this won't hold true? Adding 0s to y will only make the issue worse!
This is a very informal walkthrough of the proof, but hopefully it helps understand why it works.
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This is purely a math question!
I simply can't get my head around this formula. It is so simple, but I can't get it to work in general.
I have 2 columns of numbers.
Columns A+B is the full number of apples (100%).
Column A: is the number of apples that I have eaten.
Column B: is the number of apples that are left.
I need column C to be the percentage of apples that are left out of the full number.
For example.
A B C
0 3 100%
5 0 0%
2 2 50%
Can anybody wrap their heads around this?
Total = A + B
Left out apples = B
Percentage of left out apples = (B / Total) * 100
Consequently, here's the formula:
Column C = ((Column B) / (Column A + Column B) * 100)
Here's what you should type-in in Excel:
=(B1/(A1+B1))*100
A = No. of apples eaten, B = No. of apples left . A+B = total no of apples. So Total no. of apples left out of the full amount are (A+B)-A/100.
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I know that an affine cipher substitutes BD with SG. I need to find the encryption formula, in the form y = a x + b, where a and b are coefficients.
From the information above I end up having to equations:
a+b=18 and
3a+b=6
So I am working like this:
a+b=18 and 3a + b = 6-> 3a+18-a=6-> 2a= 6-18 -> 2a=14 (as it is mod 26)
b=18-a
2a=?
So, O want to multiply by the multiplicative inverse of 2 mod 26
I can't find a multiplicative inverse of number 2 with 26 (y = ax + b mod 26)
Can anyone please help me find a and b?
That's because 2 doesn't have a multiplicative inverse mod 26: since 13*2=0, there does not exist K such that K * a = 1. Your modulus must be prime. Try looking up the Chinese Remainder Theorem for more information.
To be more specific, integers mod 26 is not a field (a mathematical set where every element, except 0, has a multiplicative inverse). Any ring in which a * b = 0, for some a!=0 and b!=0, is not a field.
In fact, a field will always have p^n elements, where p is a prime number and n is a positive integer. The simplest fields are just integers mod a prime number, but for prime powers you need to construct a more elaborate system. So, in short, use a different modulus like 29.
Does a = 7 work? 2*7 = 14. Thus, b = 11.
Let's check the 2 equations to see if that works:
7+11 = 18 (check for the first equation).
3*7+11=21+11 = 32 = 6.
What is wrong with the above?
EDIT: Ok, now I see what could go wrong with trying to do a division by 2 in a non-prime modulus as it is similar to a division by 0. You could take ribond's suggestion of using the Chinese Remainder Theorem and split the equations into another pair of pairs:
mod 13: a+b=5, 3a+b=6. (2a = 1 = 14 => a=7. b = 18-7 = 11.)
mod 2: a+b=0. 3a+b=0 (Note this is the same equation and has a pair of possible solutions where a and b are either 0 or 1.)
Thus there is the unique solution for your problem I think.
Other posters are right in that there is no inverse of 2 modulo 26, so you can't solve 2a=14 mod 26 by multiplying through by the inverse of 2. But that doesn't mean that 2a=14 mod 26 isn't solvable.
Consider the general equation cx = d mod n (c=2,d=14,n=26 in your case). Let g = gcd(c,n). The equation cx=d has a solution if an only if g divides d. If g divides d, then there are in fact multiple solutions (g of them). The equation (c/g)x = d/g mod n/g has a unique solution (call it x_0) because c/g is relatively prime to n/g and therefore has an inverse. The solutions to the original equation are x_0, x_0 + n/g, ..., x_0 + (g-1)n/g.
In your case c=2,d=14,n=26, and g=2. g divides d, so first solve the equation (2/2)x = (14/2) mod (26/2) which gives 7. So both 7 and 7+13=20 solve your original equation.
Note that this means you haven't uniquely determined your affine transformation, two possibilities still exist. You need another data point...