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Dafna1 [17]
3 years ago
14

Please Help.....................................................

Mathematics
1 answer:
cluponka [151]3 years ago
7 0
68% due to more 1,2, and 3's
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- Two consecutive integers are 5 and 6. Write a quadratic equation that
mario62 [17]

Answer:

  x^2 -11x +30 = 0

Step-by-step explanation:

If these two integers are solutions of the quadratic, then its factors are ...

  (x -5)(x -6) = 0

Multiplying this out, we get ...

  x^2 -11x +30 = 0

6 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
A 2.5m long ladder leans against the wall of a building. The base of the ladder is 1.5m away from the wall. What is the height o
katen-ka-za [31]
I’m pretty sure you just use pythag for that so 1.5^2 + h^2 = 2.5^2 and once you solve that you should have your answer and don’t forget to root it after
7 0
2 years ago
Read 2 more answers
Red and grey bricks were used to build a decorative wall. The number of red bricks and grey bricks was 5/2. There were 140 brick
ozzi

Answer:

100 red bricks

Step-by-step explanation:

Let's say red bricks is equal to 5x

Gray bricks is equal to 2x

We have an equation

2x + 5x = 140

= 7x = 140

Divide through by 7 to get the value of x

X = 140/7

X = 20

Red bricks = 5(x)

= 5(20)

= 100

Gray bricks = 2(x)

= 2(20)

= 40

Therefore in conclusion the number of red bricks is 100.

5 0
2 years ago
Plzz i need help like rn!!!! plz :(
frozen [14]

Answer:

Step-by-step explanation:

The second choice down is the one you want.  I'm not sure why you're confused if you simply have to graph the 2 functions to see on your calculator where they intersect.  Unless you don't know how to access the change of base function in a TI84...

Hit "alpha" then "window" and 5 will open up the option to enter a base on a log.

8 0
3 years ago
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