Let A = {1, 2, 3, 4, 5, 6}, and consider the following equivalence relation R on A: R = {(1, 1),(2, 2),(3, 3),(4, 4),(5, 5),(6, 6),(2, 3),(3, 2),(4, 5),(5, 4),(4, 6),(6, 4),(5, 6),(6, 5)}. Show the partition of A defined by the equivalence classes of R.

A = C₁ U C₂ U C₄ = ,

C₁ = 'class of one' =  {1}

C₂ = 'class of two' =  {2,3}

C₄ = 'class of four' = {4,5,6}

Step-by-step explanation:

We need to find the equivalence classes of A. Since 1 does only relates to itself, then the class of 1 only has 1 element, which is 1 itself.

2, apart from itself, only does relate with 3. Thus, the equivalence class of 2 contains 2 elements: 2 and 3 (note that 3 also relates only with 2 apart fromm itself.

The remaining 3 elements, 4, 5 and 6 relate between each other, thus, they form an entire equivalence class, the equivalence class of 4.

Those are the equivalence classes of A.

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

happy to

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step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for