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docker41 [41]
3 years ago
13

Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if

a relation is not an equivalence relation, please list each property that is missing and the reason why it is missing.) (1) {(0,0), (1,1), (2,2), (3,3)} (2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)} (3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)} (4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}
Mathematics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

  • Reflexivity: If \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
  • Transitivity: \forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

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John goes from a sauna at 115 Fahrenheit to an outside temperature of -30 Fahrenheit. What is the change in temperature?
IRINA_888 [86]

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Step-by-step explanation:

7 0
2 years ago
Six distinct integers are picked from the set {1, 2, 3,…, 10}. How many selections are there, in which the second smallest integ
Ksivusya [100]

Answer:

1680 ways

Step-by-step explanation:

Total number of integers = 10

Number of integers to be selected = 6

Second smallest integer must be 3. This means the smallest integer can be either 1 or 2. So, there are 2 ways to select the smallest integer and only 1 way to select the second smallest integer.

<u>2 ways</u>   <u>1 way</u>  <u>       </u>  <u>        </u>  <u>        </u>  <u>        </u>

Each of the line represent the digit in the integer.

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Therefore, the total number of ways to form 6 distinct integers according to the given criteria will be = 1 x 2 x 840 = 1680 ways

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6 0
3 years ago
8 times 8 times 8 times 8 as a power​
Dmitrij [34]

Answer:

tex]8^{4}[/tex]

Step-by-step explanation:

8^2 = 8 x 8 = 64

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7 0
3 years ago
A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student
Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that p = 0.25

Class of 18 students

This means that n = 18

What would be the probability that at least 6 students drive themselves to school?

This is

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006

P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034

P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096

P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17

P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213

P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282

Closest option is B, just a small rounding difference.

4 0
3 years ago
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