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snow_lady [41]
3 years ago
14

18. Jameson and Quincy each deposit $7,000 into accounts that earn 6%

Mathematics
1 answer:
mina [271]3 years ago
6 0

Answer:

Jameson will have about $267.58 less in his account than Quincy

Step-by-step explanation:

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abruzzese [7]
Slope formula is y=mx+b so m is the slope and b is the y-int all u do is replace the numbers with the letters and they give u the formula which would be y=-8x-3
4 0
2 years ago
Helppppppp! will give brainliest if correct
leva [86]

Answer:

An isosceles triangle is one with 2 sides of the same length.

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7 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
Which expression represents the phrase, "half the sum of 10 and a number"? 12x+10
Nonamiya [84]
I think it is 12(10+x)
4 0
3 years ago
How many ways can Marie choose 2 pizza toppings from a menu of 6 toppings if each topping can only be chosen once?
m_a_m_a [10]

Answer:

30 different ways

Step-by-step explanation:

Hi

for the first topping she can choose 1 of 6 items

so she has 6 choices

and then the second topping she has 5 choices as each topping can only be chosen once

so it comes 6 * 5 = 30

5 0
3 years ago
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