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rodikova [14]
3 years ago
13

Let D be the ordered set of all possible words (not just English words, all strings of letters of arbitrary length) using the La

tin alphabet using only lower case letters. The order is the lexicographic order as in a dictionary (e.g. aaa < dog < door). Let A be the subset of D containing the words whose first letter is 'a' (e.g. a A, abcd ∈ A). Show that A has a supremum and find what it is.
Mathematics
1 answer:
Lapatulllka [165]3 years ago
5 0

Answer:

To show that A has a supremum is given as,

Step-by-step explanation:

Because A is that the set of all letters from the dictionary starting with the letter 'a'. And as after set A, there'll be a group containing all the opposite words .. so there exists some x during a such every other letter during a comes before it. So it's none aside from a word starting with the letter b therefore the supremum of the set A is letter b.

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Pilates is a popular set of exercises for the treatment of individuals with lower back pain. The method has six basic principles
Bas_tet [7]

Answer:

a) We reject H₀     We find difference between the mean of the two groups

b) we accept H₀ : There is no statistical difference  ( in the scale of pain) when the difference in samples means is 1

Step-by-step explanation:

Group 1: Only with educational material

Sample size     n₁  =  42

Sample mean   x₁  =  3,2

Sample standard deviation  s₁  =  2,3

Group 2: With exercises with the treatment of individuals

Sample size    n₂   =  42

Sample mean    x₂  = 5,2  

Sample standard deviation   s₂  = 2,3

Test Hypothesis

Null Hypothesis                     H₀           x₂  -  x₁  = 0  or    x₁  =  x₂

Alternative Hypothesis         Hₐ           x₂  -  x ₁ > 0  or    x₂  >  x₁

Significance level  α  = 0,01

Sample sizes are n₁  =  n₂  >  30

We use z-test

for α  = 0,01     z(c)  ≈  2,32

To calculate z(s)

z(s)  =  (  x₂  -  x₁ ) / √ (2,3)²/42  + (2,3)²/42

z(s)  = ( 5,2  -  3,2 )/ √0,2519

z(s)  =  2 / 0,5

z(s) = 4

Comparing  z(s)  and z(c)

z(s) > z(c)

z(s) is in the rejection region. We reject H₀ the true average pain level for the control condition exceeds that for the treatment condition.

b) Does the true average pain level for the control condition exceed that for the treatment condition by more than 1.

In this case

z( s) =  x₂ - x₁ / 0,5

z(s) =  1 /0,5

z(s) = 2

Comparing z(s) and z(c)  now

z(s) < z(c)        2 < 2,32

And z(s) is in the acceptance region ( for the same significance level) and we should accept H₀  equivalent to say that the two groups statistical have the same mean

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Answer:

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