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We are in the process of planning our first sprint using scrum poker cards and we were wondering,
why are there missing numbers in scrum poker (i.e. 4,6,...)?
One commercially available set of Scrum poker cards use a sequence similar to the Fibonacci sequence:
0, ½, 1, 2, 3, 5, 8, 13, 20, 40, 100, and optionally a ? (unsure) and
a coffee cup (I need a break)
As for why, it's to emphasise that difficulty does not increase linearly, it usually increases exponentially (hence the approximation of Fibonacci sequence).
Also see this post for other ideas as to the choice of this sequence.
The sequence of numbers used for prioritizing in Scrum is a Fibonacci sequence. The reason the Fibonacci sequence is that experience shows that the sequence tends to be accurate for the way that people estimate difficulty/level of effort.
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Need some help with a homework question.
I suck at math and our prof does not show us how to break down equations into R code at all, nor does she discuss R at all, but expects us to use it. Can someone show me how to work out this problem in R? A classmate and myself have been working on homework all day and we're exhausted and I've got a migraine setting in. Any help is much appreciated!
A population consists of N=10 primary units, each of which consists of Mi=6 secondary units. A two-stage sampling design selects 2 primary units by simple random sampling (without replacement) and 3 secondary units from each selected primary unit, also by simple random sampling. The observed values of the variable of interest are 7, 5, 3 from the first primary unit selected and 4, 2, 3 from the second primary unit selected.
The population mean per secondary unit was 4.
Estimate the variance of the estimator above.
Not sure where to start
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I want to understand the physical significance of n raise to some decimal power.
Like when i say 2^5. I understand that it means 2 multiplied 5 times. But how do i analyse 2^0.1.
Please suggest.
2^0.1 is the tenth root of 2. For rational powers, x^(p/q)=(x^p)^(1/q) is a combination of powers and roots.
For general real numbers,
x^y = exp(log(x)*y).
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The formula for the calculation of the amount of humans per generation is 2^n
n 2^n
0 1 that's me
1 2 this is the count of my parents (father + mother)
2 4 this is the count of my grand-parents
and so on
Assuming that every generation creates the next one with the age of 20 then we have in 2000 years 100 generations, that means:
n 2^n
100 1.267.650.600.228.229.401.496.703.205.376 humans has been living on earth!
And now I am reading in wikipedia: The population on earth is today estimated 7.3 billion.
Comparing these to numbers is amazing:
1.267.650.600.228.229.401.496.703.205.376
7.600.000.000
Is this possible or did I make a mistake?
Your calculations seem to be correct but you are missing some important factors.
The most critical miss is that you are assuming that all these ancestors must be distinct – that one person cannot occupy two different places in your family tree.
In most families you will find fairly recent examples of cousins marrying each other, especially if you use the wider definition of cousin that isn't restricted to First Cousins.
This significantly reduces the number of ancestors in your tree.
There is a saying that, after enough generations, either everybody on Earth will be your descendent, or else nobody will be, and there is a lot of truth in this.
We are all one big family when you think of it that way!
This is a well-known phenomenon, and is referred to as Pedigree Collapse
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I am trying to calculate the computational complexity of ifft, I know it’s NlogN for a N*1 dimension, signal. But I have a product of two signals then I want to get ifft, and then calculate the computational complexity. So simply, if X(w) and Q(w) are Fourier transform of two time signal , then what is computational complexity of their product.
Note: X(w) and Q(w) has same (N*1) size.
ifft(X(w)*Q(w))=???
It is still O(N log N). The ifft doesn't care how you get the data, and the element-wise multiplication is O(N).
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Suppose you have a deck of 100 cards, with the numbers 1-100 on one side. You select a card, note the number, replace the card, shuffle, and repeat.
Question #1: How many cards (on average) must you select to have drawn the same card twice? Why?
Question #2: How many cards (on average) must you select to have drawn all of the cards at least once? Why?
(thanks, it has to do with random music playlists and making the option to not repeat the shuffle, as it were)
Q1: Relates to Birthday paradox problem
As you see in the collision problem section(in wikipedia link above), your question maps exactly.
Cast as a collision problem
The birthday problem can be generalized as follows: given n random integers drawn from a discrete uniform distribution with range [1,d], what is the probability p(n;d) that at least two numbers are the same? (d=365 gives the usual birthday problem.)
You have a range [1,100] from which you select random cards. The probability of collision(two selected cards are the same) is given as p(n;d) = ...
Further down, we have formula for average/expected number of selections as
Q(100) gives your answer.