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Anna71 [15]
3 years ago
15

A water tank with 265 gallons of water has a leak. After 5 hours, the tank is completely empty.

Mathematics
2 answers:
ddd [48]3 years ago
3 0
<span>The water tank loses 53 gallons of water in 1 hour.</span>
Shtirlitz [24]3 years ago
3 0
265÷5=53, so the water tank loses 5e gallons of water in 1 hour
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50 points and a BRAINLIEST if you answer correctly! Pls Pls help!
Gre4nikov [31]

Answer:

3/10

Step-by-step explanation:

Numbers Given are:

1 1/2, 2 1/2, 1 2/3, and 2 5/6

The result at the end of all the math is 2 5/6.

1 2/3 + 2 5/6 = 4 1/2

Divide 4 1/2 by 1 2/3 to get 2 7/10.

Subtract 2 2/3 from 2 7/10 to get 1/5.

Multiply 1/5 by 1 1/2 to get 3/10.

So the number they thought of was 3/10.

4 0
2 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
Which is a perfect square ?
barxatty [35]

Answer:

6^2 Is perfect square which is 36.

3 0
2 years ago
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3241004551 [841]

Use angle sum property

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<A=68

<B=44

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2 years ago
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Find the area of the shaded region.<br>pls do this with explanation.<br>​
vladimir1956 [14]

Answer:

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6 0
2 years ago
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