Answer:
For question a, it simplifies. If you re-express it in boolean algebra, you get:
(a + b) + (!a + b)
= a + !a + b
= b
So you can simplify that circuit to just:
x = 1 if b = 1
(edit: or rather, x = b)
For question b, let's try it:
(!a!b)(!b + c)
= !a!b + !a!bc
= !a!b(1 + c)
= !a!b
So that one can be simplified to
a = 0 and b = 0
I have no good means of drawing them here, but hopefully the simplification helped!
Answer:
By Using the Greedy- Activity- Selection algorithm
Explanation:
The Greedy- Activity- Selection algorithm in this case involves
First finding a maximum size set S1, of compatible activities from S for the first lecture hall.
Then using it again to find a maximum size set S2 of compatible activities from S - S1 for the second hall.
This is repeated till all the activities are assigned.
It requires θ(n2) time in its worse .
The answer is the team. Teams are self-establishing which no one not even the scrum master tells the team how to turn product backlog into augmentations of stoppable functionality. Each team member put on his or her knowledge to all of the complications. The interaction that outcomes progresses the entire scrum team’s general competence and usefulness. The optimal size for a scrum team is seven people, plus or minus two. When there are fewer than five team members, there is less interface and as a result less productivity improvement. The product owner and scrum master roles are not comprised in this count. Team arrangement may variation at the end of a sprint. Every time Team membership is altered, the productivity increased from self-organization is reduced.
