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Feliz [49]
3 years ago
14

Show that the relation R consisting of all pairs(x, y)such that x and y are bit strings of length three or more that agree in th

eir first three bits is an equivalence relation on the set of all bit strings of length three or more
Mathematics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

See proof below

Step-by-step explanation:

An equivalence relation R satisfies

  • Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
  • Symmetry: For all x,y, if xRy then yRx.
  • Transitivity: For all x,y,z, If xRy and yRz then xRz.

Let's check these properties: Let x,y,z be bit strings of length three or more

The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.

If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.

If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.

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mihalych1998 [28]

Answer:

No, it is not okay to conduct the simulation this​ way.

Step-by-step explanation:

In statistics, simulation refers to a technique that is employed to model random events so that the results obtained from using the simulation is significantly similar to the results obtained from observing the real-world.

Researchers are therefore able to understand the real world when they observe the simulated outcomes.

From the description above, it can be seen that simulation is about studying random events. Therefore, a sample of the population that will be used in the simulation must be selected through a random sampling.

Random sampling refers to the sampling method that gives equal opportunity of being selected to each member of the population. This makes the sample selected through random sampling technique to be an unbiased representation of the total population.

As a result, making up 31 numbers between 1 and 365 by the student is not a random sampling, because his method may favor some numbers over others. It is therefore a defective method of carrying out simulation.

 Therefore, the it is not okay to conduct the simulation this​ way.

I wish you the best.

6 0
3 years ago
Simplify negative 4 and 1 over 4 − negative 9 and 1 over 2.
Lorico [155]
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8 0
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kari74 [83]
The measurements of an angle X is 80
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3 years ago
One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r
Flura [38]

Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

3 0
3 years ago
Read 2 more answers
A sequence is defined by the recursive function f(n + 1) = one-halff(n). If f(3) = 9 , what is f(1) ?
almond37 [142]

Answer:

d.) 81

Step-by-step explanation:

f(2) would be 27 and 27x3 is 81.

81/3=27

27/3=9

9 0
4 years ago
Read 2 more answers
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