<![CDATA[Math Is Fun Forum / Help Me !]]> 2019-05-21T06:45:02Z FluxBB http://www.mathisfunforum.com/index.php <![CDATA[Reviewing area of polygons]]> Q5 has recently and many times earlier been answered in other posts.  Use the search to find these.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2019-05-21T06:45:02Z http://www.mathisfunforum.com/viewtopic.php?id=24974&action=new
<![CDATA[Probability Distribution]]> hi Zeeshan 01

The question doesn't make clear if you are expected to fit a formula to this data, such as Poisson.  Also it says nothing about the time scale under which the data was collected. If, for example, we had been told " 'x' shows the number of machine breakdowns during each month collected over the last several years" , we could then estimate the probability of getting more than 2 breakdowns in the next week.

As we're not told this and not asked for such answers, I'd take the simplest interpretation and just calculate total f = T and the probability that X > n by adding a third row to the table.  Then for example P(X>6) = (4 + 3 + 1)/T

[X only takes discrete values so a continuous function is not appropriate I think]

Bob

If you have been given an example of a question like this, then post it and it may make things clearer.

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http://www.mathisfunforum.com/profile.php?id=67694 2019-05-20T07:22:25Z http://www.mathisfunforum.com/viewtopic.php?id=24968&action=new
<![CDATA[Sequence]]> It is clear that none of the terms in the sequence is divisible by 5. For

Similarly none of the terms is divisible by 3, for

It follows that

i.e. consecutive terms are coprime; also

i.e. alternate terms are coprime as well.

Since a₁ = 1, k =1 fits the bill. And since the sequence is clearly strictly increasing, it is the only one that does.

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-19T14:39:03Z http://www.mathisfunforum.com/viewtopic.php?id=8161&action=new
<![CDATA[A linear algebra problem]]>

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-19T11:04:48Z http://www.mathisfunforum.com/viewtopic.php?id=1959&action=new
<![CDATA[topology help]]> https://en.wikipedia.org/wiki/Ham_sandwich_theorem.

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-18T20:45:24Z http://www.mathisfunforum.com/viewtopic.php?id=9971&action=new
<![CDATA[inequality]]> Try Muirhead’s inequality and the fact that the sequence (4,0,0) majorizes (3,1,0) (i.e 4 ≥ 3, 4+0 ≥ 3+1, 4+0+0 ≥ 3+1+0).

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-18T19:10:54Z http://www.mathisfunforum.com/viewtopic.php?id=12893&action=new
<![CDATA[abstract algebra]]> Let G be a group of order 595 = 5 × 7 × 17.

The number n of Sylow 17-subgroups divides 5 × 7 = 35 and is ≡ 1 (mod 17). Possibilities are 1 and 35.

Suppose 35. Then the intersection of all these subgroups (any two of which only have the identity in common since 17 is prime) would contain 1 + 35×(17−1) = 561 elements, i.e. 560 elements of order 17. Now, if P is a Sylow 17-subgroup and Q is a Sylow 5-subgroup, then PQ would be a subgroup of order |P||Q|/|PQ| = 85; moreover it would be cyclic, since P and Q are cyclic and their orders are coprime. But a cyclic subgroup of order 85 has φ(85) = 64 generators, i.e. 64 elements of order 85 – and we already have 560 elements of order 17.

Hence there can only be 1 Sylow 17-subgroup.

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-18T15:35:25Z http://www.mathisfunforum.com/viewtopic.php?id=14522&action=new
<![CDATA[Use Extreme Value Theorem]]>

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-18T02:21:26Z http://www.mathisfunforum.com/viewtopic.php?id=14523&action=new
<![CDATA[I need help in Algebra (Sylow Theorems)!]]>

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-18T00:51:03Z http://www.mathisfunforum.com/viewtopic.php?id=14531&action=new
<![CDATA[logic]]> I would use:

• Hxy : x hides y in public,
hxy : x has y,
Mx : x is modest,
a : Angus,
p : private bits.

Then the premises are:

• x((Mx∧hxp)→Hxp),
y(hay→¬Hay).

The first premise is equivalent to

•∀x(¬Hxp→¬(Mx∧hxp)) ≡ ∀x(¬Hxp→(¬Mx∨¬hxp)).

Substituting x = a into it gives

•¬Hap→(¬Ma∨¬hap).

Substituing y = p into the second premise gives

•hap→¬Hap.

It follows that

•hap→(¬Ma∨¬hap) ≡ ¬hap∨(¬Ma∨¬hap) ≡ ¬Ma∨¬hap

which is the conclusion. Therefore the argument is valid.

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-17T23:52:38Z http://www.mathisfunforum.com/viewtopic.php?id=14556&action=new
<![CDATA[Find the order of the cyclic subgroup of D2n generated by r]]> xsw001 wrote:

Does it implies that the order of cyclic subgroup {1, r, r^2, ... , r^n-1} of D2n is n then?

Yes.

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-17T22:17:13Z http://www.mathisfunforum.com/viewtopic.php?id=15024&action=new
<![CDATA[Boundary points]]> Try

• .

Then

and

so

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-17T20:04:02Z http://www.mathisfunforum.com/viewtopic.php?id=15074&action=new
<![CDATA[proof of same cardinality]]> Note that

is a partition of A since f is surjective and (as a bona fide function) is well defined. So, use the axiom of choice to pick for each bB exactly one aᵇf⁻¹(b). This will give you an injection

• .

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-17T14:33:16Z http://www.mathisfunforum.com/viewtopic.php?id=15223&action=new
<![CDATA[Intergration Problem Please Help]]> If you shift the curve 2 units to the right, you’ll be rotating the curve y = √(4−[x−2]) around the y-axis instead.

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http://www.mathisfunforum.com/profile.php?id=217821 2019-05-17T13:27:27Z http://www.mathisfunforum.com/viewtopic.php?id=15242&action=new
<![CDATA[Euler Characteristic - is it true that...]]> Thanks - the text I was reading probably did mean 'homeomorphic', so you've given me the answer.

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http://www.mathisfunforum.com/profile.php?id=212622 2019-05-17T12:53:00Z http://www.mathisfunforum.com/viewtopic.php?id=23245&action=new