Answer:
even degree with positive leading coefficient for end behavior x ----> infinity, y ---> infinity
Step-by-step explanation:
I have a chart you can use
Answer:
y=525x+230
Step-by-step explanation:
525 is spent everyday. x is the number of days, so with each day $525 is spent.
$230 is a one time cost, regardless of how many days they stay on the trip.
160 i believe. to find the area of a triangle act like its a square and divide in half. for squares multiply base by height.
take the area of each shape and add it up
The Karger's algorithm relates to graph theory where G=(V,E) is an undirected graph with |E| edges and |V| vertices. The objective is to find the minimum number of cuts in edges in order to separate G into two disjoint graphs. The algorithm is randomized and will, in some cases, give the minimum number of cuts. The more number of trials, the higher probability that the minimum number of cuts will be obtained.
The Karger's algorithm will succeed in finding the minimum cut if every edge contraction does not involve any of the edge set C of the minimum cut.
The probability of success, i.e. obtaining the minimum cut, can be shown to be ≥ 2/(n(n-1))=1/C(n,2), which roughly equals 2/n^2 given in the question.Given: EACH randomized trial using the Karger's algorithm has a success rate of P(success,1) ≥ 2/n^2.
This means that the probability of failure is P(F,1) ≤ (1-2/n^2) for each single trial.
We need to estimate the number of trials, t, such that the probability that all t trials fail is less than 1/n.
Using the multiplication rule in probability theory, this can be expressed as
P(F,t)= (1-2/n^2)^t < 1/n
We will use a tool derived from calculus that
Lim (1-1/x)^x as x->infinity = 1/e, and
(1-1/x)^x < 1/e for x finite.
Setting t=(1/2)n^2 trials, we have
P(F,n^2) = (1-2/n^2)^((1/2)n^2) < 1/e
Finally, if we set t=(1/2)n^2*log(n), [log(n) is log_e(n)]
P(F,(1/2)n^2*log(n))
= (P(F,(1/2)n^2))^log(n)
< (1/e)^log(n)
= 1/(e^log(n))
= 1/n
Therefore, the minimum number of trials, t, such that P(F,t)< 1/n is t=(1/2)(n^2)*log(n) [note: log(n) is natural log]
