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Cloud [144]
3 years ago
13

How are polynomials like other number systems such as whole numbers and integers?

Mathematics
1 answer:
Agata [3.3K]3 years ago
8 0
You can add, subtract, and multiply them. These three operations obey the rules for integers. There's a polynomial division algorithm that fills formally the same role as the usual division algorithm for integers. Polynomials added to, subtracted from, or multiplied by other polynomials yield only polynomials. Likewise, integers added to, subtracted from, or multiplied by other integers yield only integers.
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A = 3 and c = 12 c/a =
Bess [88]

Answer:

4, because 12/3 is 4 cause 3×4=12

3 0
2 years ago
From a group of 11 people, 4 are randomly selected. What is the probability the 4 oldest people in the group were selected
asambeis [7]

Answer:

<h2>4/11</h2>

Step-by-step explanation:

Probability is the likelihood or chance that an event will occur.

<em>Probability = expected number of outcome/Total number of outcome</em>

If there are group of 11 people, then the total outcome of events will be 11

If we are to select 4 oldest people from the group, then the expected outcome is 4.

Hence, the probability that the 4 oldest people in the group were selected is 4/11.

3 0
3 years ago
• karger's min cut algorithm in the class has probability at least 2/n2 of returning a min-cut. how many times do you have to re
MrRissso [65]
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]
4 0
3 years ago
Complete the inequality statement. 4 1/4___ 17/4
Mumz [18]

Answer:

they are =

im sorry it was a little late

8 0
3 years ago
Consider this function rule: multiply the input by two and then subtract one to get the output. Write an equation that gives thi
rusak2 [61]

Answer:

f(x)=2x-1

Step-by-step explanation:

The output is f(x). The input is x.

The prompt says to multiply the input by 2, so 2x.

Then subtract 1 to get f(x).

Hope this helps!

7 0
2 years ago
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