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Zina [86]
2 years ago
13

A farmer can plant up to 6 acres of land with soybeans and corn. Her use of a necessary pesticide is limited by federal regulati

ons to 15 gallons for her entire 6 acres. Soybeans require 2 gallons of pesticide for every acre planted and corn requires 3 gallons per acre. The profit the farmer makes by earning $4,000 for every acre of soybeans he plants and $3,000 for every acre he plants with barley can be modeled by P=4000x+3000y . If x represents acres of soybeans and y represents acres of corns, which inequalities represent the possible solutions to her situation?
Mathematics
1 answer:
horrorfan [7]2 years ago
5 0
The answer is x≥0 y≥0 x+y≤6 2x+3y≤15
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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
In the diagram, point O is the center of the circle and mADB = 43°. If mAOB = mBOC, what is mBDC?
Vanyuwa [196]
The answer is B) 43<span>°
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7 0
3 years ago
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docker41 [41]
It’s the 1st one 5,5
5 0
2 years ago
Please answer this question right now ​
Westkost [7]

Answer:

6) c

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Hope this helps!

6 0
3 years ago
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Answer:

904.778 in.

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4 0
2 years ago
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