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max2010maxim [7]
3 years ago
13

You are randomly picking a number between 1 and 20, inclusive. What is the probability of selecting a

Mathematics
2 answers:
julsineya [31]3 years ago
5 0

Answer:

of favourable events = 4, (which includes 1, 4, 9, 16). And, total number of observations are 20. So, the probability of picking a square number between 1 and 20 is 1/5.

Step-by-step explanation:

andrey2020 [161]3 years ago
4 0

Answer:

Answer is 20% chance. (20/100 or 4/20)

Step-by-step explanation:

There are 4 perfect squares in a number set 1-20.

1,4,9, and 16.

So that is 4/20. I multiplied the fraction by 5 to get 20/100 to get 20%

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Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.
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Step-by-step explanation:

A = {0,1,2,3}

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b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric, because if the (a,b)∈R and if the (b, a) ∈ R, than a=b (since (2,0) ∈ R and (0,2) ∉ R; and (2,3) ∈ R and (3,2) ∉ R )

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R is transitive , because if the (a,b)∈R and if the ( b , c )∈R . then a = b or b = c ( since there are only two element not of the form ( a , a ) and that pair does not satisfy ( a,b ) ∈ R and ( b , a ) ∈ R ), which implies ( a , c ) = ( b , c ) ∈ R or ( a , c ) = ( a , b ) ∈ R.

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c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive, because (a,a)∈R of every element a ∈ A.

R is antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R . then a = b ( since ( 1 , 2 )∈R and ( 2 , 1 ) ∉ R; ( 3 , 1 ) ∈ R and ( 1 , 3 ) ∉ R ).  

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R is not a partial ordering. because R is not transitive .

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

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R is the antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R, then a = b ( since ( 1 . 2 )∈R and ( 2 . 1 )∉R; similarly, all other elements not of the form (a,a) ).

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R is not a partial ordering, because R is not the antisymmetric and not the transitive.

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