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Mathematical Ramblings

Tuesday, September 25, 2007. Looks like I've had this wrong in my head for a while now. Let k be the percentile rank in question, divided by 100, let S be the data set in question, and let N be the number of readings in S, i.e. N = #(S). Instead, we need to refer to the exact definition of P(k), which is the number below which exactly (k*100)% of the readings fall. For a continuous application such as ours, Vic G suggests we consider the matter in the following way:. 20 = 11.1.%. 30 = 22.2.%. The 90th pe...

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Mathematical Ramblings | mathematique.blogspot.com Reviews
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Tuesday, September 25, 2007. Looks like I've had this wrong in my head for a while now. Let k be the percentile rank in question, divided by 100, let S be the data set in question, and let N be the number of readings in S, i.e. N = #(S). Instead, we need to refer to the exact definition of P(k), which is the number below which exactly (k*100)% of the readings fall. For a continuous application such as ours, Vic G suggests we consider the matter in the following way:. 20 = 11.1.%. 30 = 22.2.%. The 90th pe...
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1 mathematical ramblings
2 calculating percentile ranks
3 labels statistics
4 1 comments
5 the chomsky hierarchy
6 unrestricted
7 context sensitive
8 labels formal grammars
9 0 comments
10 either
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mathematical ramblings,calculating percentile ranks,labels statistics,1 comments,the chomsky hierarchy,unrestricted,context sensitive,labels formal grammars,0 comments,either,c = h,a = h,isin h,and so,b = h,h = {h,hg = {h,where h,g &isin g,g = h,proof
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Mathematical Ramblings | mathematique.blogspot.com Reviews

https://mathematique.blogspot.com

Tuesday, September 25, 2007. Looks like I've had this wrong in my head for a while now. Let k be the percentile rank in question, divided by 100, let S be the data set in question, and let N be the number of readings in S, i.e. N = #(S). Instead, we need to refer to the exact definition of P(k), which is the number below which exactly (k*100)% of the readings fall. For a continuous application such as ours, Vic G suggests we consider the matter in the following way:. 20 = 11.1.%. 30 = 22.2.%. The 90th pe...

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mathematique.blogspot.com mathematique.blogspot.com
1

Mathematical Ramblings: Calculating Percentile Ranks

http://www.mathematique.blogspot.com/2007/09/calculating-percentile-ranks.html

Tuesday, September 25, 2007. Looks like I've had this wrong in my head for a while now. Let k be the percentile rank in question, divided by 100, let S be the data set in question, and let N be the number of readings in S, i.e. N = #(S). Instead, we need to refer to the exact definition of P(k), which is the number below which exactly (k*100)% of the readings fall. For a continuous application such as ours, Vic G suggests we consider the matter in the following way:. 20 = 11.1.%. 30 = 22.2.%. The 90th pe...

2

Mathematical Ramblings: A closed subset of a group is a group

http://www.mathematique.blogspot.com/2006/06/closed-subset-of-group-is-group.html

Friday, June 02, 2006. A closed subset of a group is a group. Let (G, ⋅) be a group, and let H ⊂ G. If H is closed under ⋅, then H is a subgroup. Let H have m elements, so that. By the cancellation laws, Ha. Has m elements as well, and since H is closed and a. Isin; H, Ha. Isin H = Ha. Isin H such that a. E (∈H). (1). Also, ∃a. Isin H such that a. That is, a. Similarly, by looking at Ha. For i = 2.m, a. Isin H. (2). Has identity (1), and. Therefore, H is a group, and so H is a subgroup of G.

3

Mathematical Ramblings: Introductory Post

http://www.mathematique.blogspot.com/2006/05/introductory-post.html

Sunday, May 28, 2006. One day, I'll write some maths stuff here. Posted by Escheresque @ 3:07 am. And read the other 75% of books on your shelf? This blog was supposed to be my secret maths mutterings! I will never read the other 75% of my books, because I buy books much faster than I can read them!

4

Mathematical Ramblings: September 2007

http://www.mathematique.blogspot.com/2007_09_01_archive.html

Tuesday, September 25, 2007. Looks like I've had this wrong in my head for a while now. Let k be the percentile rank in question, divided by 100, let S be the data set in question, and let N be the number of readings in S, i.e. N = #(S). Instead, we need to refer to the exact definition of P(k), which is the number below which exactly (k*100)% of the readings fall. For a continuous application such as ours, Vic G suggests we consider the matter in the following way:. 20 = 11.1.%. 30 = 22.2.%. The 90th pe...

5

Mathematical Ramblings: Equivalence classes are either identical or disjoint

http://www.mathematique.blogspot.com/2006/06/equivalence-classes-are-either.html

Friday, June 02, 2006. Equivalence classes are either identical or disjoint. Let S be a non-empty set with equivalence relation , and a S let. A] = {x S a x}. Be the equivalence class. Then two equivalence classes [a] and [b] are either identical or disjoint. Consider [a] [b]. If [a] [b] = , well and good. Otherwise, c S such that c [a] [b]. That is, c a and c b. By transitivity of equivalence relations, a b. If x [a] then x [b], so [a] [b]. Similarly, and by reflexivity of equivalence relations, [b] [a].

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Upside down, Miss Jane!: By the way

http://escheresque.blogspot.com/2007/09/by-way.html

Upside down, Miss Jane! Tuesday, September 25, 2007. Just to remind anyone who's potentially interested, I have two other blogs that are less what-I-had-for-lunch:. My random maths mutterings; and. I'm a Linguistics Undergraduate. Which is a collaboration between myself and some other linguistics undergrads. Posted by Escheresque at 12:37 am. I had pasta salad and chickpeas. but i don't think that was the point. No it wasn't, and yes you would get in trouble! A word of warning. Its not my fault!

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Mathematical Ramblings

Tuesday, September 25, 2007. Looks like I've had this wrong in my head for a while now. Let k be the percentile rank in question, divided by 100, let S be the data set in question, and let N be the number of readings in S, i.e. N = #(S). Instead, we need to refer to the exact definition of P(k), which is the number below which exactly (k*100)% of the readings fall. For a continuous application such as ours, Vic G suggests we consider the matter in the following way:. 20 = 11.1.%. 30 = 22.2.%. The 90th pe...

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