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mcdonalds: now for black people! (pg. 6)
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| The17sss |
| quote: | Originally posted by MrJiveBoJingles
Not all affirmative action policies mandate percentages. Affirmative action may simply be allowing employers to consider minority origin in hiring. |
if by allowing, you mean forcing... then yes A.A. may indeed allow employers to consider minorities when hiring. :p |
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| inconspicuous |
| quote: | Originally posted by chimera66
that i can agree with
do you honestly believe they hire under-qualified people just because of affirmative action? |
I was including college admissions in that, as well, and yes in both cases. If you're unfamiliar with the University of Michigan's past on the subject, you would do well to look into it. For instance, getting a 1600 on your SATs as a white male had, until recently, been worth less than getting a 100 & being black in their formula for selecting applicants for admission. no, I did not forget a zero.
| quote: | Originally posted by chimera66
it is meant to get minorities, which does not translate just to black people opportunities but they have to be qualifed for the opportunity. how many under-qualified people with connections get jobs they shouldn't? i can think if plenty of people who have jobs/opportunties they shouldn't because of an assortment of things. |
I'm against anyone getting opportunities they shouldn't. Affirmative action just compounds the number of people who do, though, exponentially.
If I was really a bastard, I'd point out that at least in the case of connections, it is an asset acquired after birth.
| quote: | Originally posted by chimera66
anyhow, what pisses me off about people talking about affirmative action is they instantly think it is all about helping black people when honestly it's to help a wide group of people including women who probably benefit more so than any other group. the only group who doesn't benefit from it are white males. is that fair, no but don't point the finger at black people specifically as if we are the only ones to benefit. |
The fact is, though, black people outnumber all non-latino minorities in the US COMBINED (blacks and latinos are almost equal in number...fractions of a percentage) and there isn't a single person on earth who is going to claim that latinos get anywhere near the same benefit from affirmative action, in sum or case-by-case.
the 'they took candy, too!' argument is totally worthless. |
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| chimera66 |
| quote: | Originally posted by The17sss
Yes... for sure. When you have to fill a certain quota that doesn't rely on merit/qualification first, by definition you're trending the wrong way... in terms of job performance.
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understandably but given the number of people who apply to one job, the chances of one minority (black, hispanic, asian, female,etc) not being qualified for the job is pretty low. i understand where you are coming from but honestly i don't believe that many under qualified people get jobs.
as for the mention of u michigan; i know it happens in colleges and i definitely don't believe it should be used that way. if anything it just shows that minorities should be held to a lower standard which perpetuates stereotypes.
| quote: | Originally posted by inconspicuous
I'm against anyone getting opportunities they shouldn't. Affirmative action just compounds the number of people who do, though, exponentially.
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that is true, i'm with you on that one.
when i mentioned connections i meant people who have their parents or other relatives get them jobs. i can't stand that because it's people with money keeping it within their circle. i know my fair share of pampered people with trust funds and all that and lots of them got jobs/internships/etc because of who they were. those type of connections shouldn't get you jobs. |
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| winston |
The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure.
For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function : G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H).
When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that
f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s)
for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth.
The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism : A → B is a function between two algebraic structures of the same type such that
f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\,
for each n-ary operation μ and for all elements a1,...,an ∈ A.
The real numbers are a ring, having both addition and multiplication. The set of all 2 Χ 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by
f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix}
where r is a real number. Then is a homomorphism of rings, since preserves both addition:
f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s)
and multiplication:
f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s).
For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function from the nonzero complex numbers to the nonzero real numbers by
f(z) = |z|.\,\!
That is, (z) is the absolute value (or modulus) of the complex number z. Then is a homomorphism of groups, since it preserves multiplication:
f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2).
Note that cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition:
|z_1 + z_2| \ne |z_1| + |z_2|. |
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| bas |
| quote: | Originally posted by winston
The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure.
For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function : G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H).
When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that
f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s)
for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth.
The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism : A → B is a function between two algebraic structures of the same type such that
f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\,
for each n-ary operation μ and for all elements a1,...,an ∈ A.
The real numbers are a ring, having both addition and multiplication. The set of all 2 Χ 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by
f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix}
where r is a real number. Then is a homomorphism of rings, since preserves both addition:
f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s)
and multiplication:
f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s).
For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function from the nonzero complex numbers to the nonzero real numbers by
f(z) = |z|.\,\!
That is, (z) is the absolute value (or modulus) of the complex number z. Then is a homomorphism of groups, since it preserves multiplication:
f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2).
Note that cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition:
|z_1 + z_2| \ne |z_1| + |z_2|. |
Ah yeah. Now it makes sense. |
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| Lira |
| quote: | Originally posted by winston
The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure.
For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function : G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H).
When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that
f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s)
for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth.
The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism : A → B is a function between two algebraic structures of the same type such that
f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\,
for each n-ary operation μ and for all elements a1,...,an ∈ A.
The real numbers are a ring, having both addition and multiplication. The set of all 2 Χ 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by
f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix}
where r is a real number. Then is a homomorphism of rings, since preserves both addition:
f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s)
and multiplication:
f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s).
For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function from the nonzero complex numbers to the nonzero real numbers by
f(z) = |z|.\,\!
That is, (z) is the absolute value (or modulus) of the complex number z. Then is a homomorphism of groups, since it preserves multiplication:
f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2).
Note that cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition:
|z_1 + z_2| \ne |z_1| + |z_2|. |
Bed + You = Advisable |
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| Spam |
| quote: | Originally posted by winston
The common theme is that a homomorphism is a function between two algebraic objects that respects the algebraic structure.
For example, a group is an algebraic object consisting of a set together with a single binary operation, satisfying certain axioms. If G and H are groups, a homomorphism from G to H is a function : G → H such that f(g_1 * g_2) = f(g_1) * f(g_2)\,\! for any elements g1, g2 ∈ G, where ∗ denotes the respective binary operations (the first ∗ denoting the operation in G, and the second ∗ denoting the operation in H).
When an algebraic structure includes more than one operation, homomorphisms are required to preserve each operation. For example, a ring possesses both addition and multiplication, and a homomorphism between two rings is a function such that
f(r+s) = f(r) + f(s)\qquad\text{and}\qquad f(rs) = f(r)\,f(s)
for any elements r and s of the domain ring. In most contexts, a homomorphism will map identity elements to identity elements, inverse elements to inverse elements, and so forth.
The notion of a homomorphism can be given a formal definition in the context of universal algebra, a field which studies ideas common to all algebraic structures. In this setting, a homomorphism : A → B is a function between two algebraic structures of the same type such that
f(\mu_A(a_1, \ldots, a_n)) = \mu_B(f(a_1), \ldots, f(a_n))\,
for each n-ary operation μ and for all elements a1,...,an ∈ A.
The real numbers are a ring, having both addition and multiplication. The set of all 2 Χ 2 matrices is also a ring, using matrix addition and matrix multiplication. Define a function between these rings by
f(r) = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix}
where r is a real number. Then is a homomorphism of rings, since preserves both addition:
f(r+s) = \begin{pmatrix} r+s & 0 \\ 0 & r+s \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} + \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r) + f(s)
and multiplication:
f(rs) = \begin{pmatrix} rs & 0 \\ 0 & rs \end{pmatrix} = \begin{pmatrix} r & 0 \\ 0 & r \end{pmatrix} \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix} = f(r)\,f(s).
For another example, the nonzero complex numbers form a group under the operation of multiplication, as do the nonzero real numbers. (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Define a function from the nonzero complex numbers to the nonzero real numbers by
f(z) = |z|.\,\!
That is, (z) is the absolute value (or modulus) of the complex number z. Then is a homomorphism of groups, since it preserves multiplication:
f(z_1 z_2) = |z_1 z_2| = |z_1|\,|z_2| = f(z_1)\,f(z_2).
Note that cannot be extended to a homomorphism of rings (from the complex numbers to the real numbers), since it does not preserve addition:
|z_1 + z_2| \ne |z_1| + |z_2|. |
Thank god someone's finally explained this to me in plain english. I don't know what I'd do without crackheads like winston explaining the world to me. |
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| winston |
| quote: | Originally posted by Spam
Thank god someone's finally explained this to me in plain english. I don't know what I'd do without crackheads like winston explaining the world to me. |
Du siehst den Wald vor lauter Bδumen nicht. |
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| Spam |
| quote: | Originally posted by winston
Du siehst den Wald vor lauter Bδumen nicht. |
No, YOU go yourself. |
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| winston |
| homme mort ne fait guerre |
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