The probability of rolling 1 number, a four, on a 6-sided number cube, is 1/6.
On the second roll, the probability of rolling 1 number, a four, is again 1/6 (remember theoretical probability for one event does not consider previous rolls).
But the probability of rolling a four on 2 on consecutive rolls will be the probability of the first event times the probability of the second event. Think of it as rolling a four on the first time <em>and</em> the second time, and whenever there is "and" you need to multiply the probabilities. The probability of rolling a four on two consecutive rolls is
(1/6)*(1/6) = 1/36.
OK I will answer give me a few minutes I’ll try to be fast and come back to you in the comment section
(1) [6pts] Let R be the relation {(0, 1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0)} defined on the set {0, 1, 2, 3}. Find the foll
goldenfox [79]
Answer:
Following are the solution to the given points:
Step-by-step explanation:
In point 1:
The Reflexive closure:
Relationship R reflexive closure becomes achieved with both the addition(a,a) to R Therefore, (a,a) is 
Thus, the reflexive closure: 
In point 2:
The Symmetric closure:
R relation symmetrically closes by adding(b,a) to R for each (a,b) of R Therefore, here (b,a) is:
Thus, the Symmetrical closure:
Answer:
BE is congruent to ED
Step-by-step explanation:
