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deff fn [24]
4 years ago
13

Let f:a to b be a surjective map of sets prove that the relation a if and only if f(a) = f(b) is an equivalence relation whose e

quivalence classes are the fibers of f
Mathematics
1 answer:
MrRa [10]4 years ago
6 0

Answer:

Step-by-step explanation:

Denote this equivalence relation by "\sim" (i.e, a\sim b if and only if f(a)=f(b)), is clear that a\sim a is an equivalence relation since "=" is. Now, by definition we have that [a]=\{b\sim a \mid f(a)=f(b) \}=f^{-1}(a).

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Answer:

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Step-by-step explanation:

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P(\bar X

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*Use a <em>z</em>-table.

Compute the value of the mean amount invested as follows:

\bar x=\mu_{\bar x}+z\cdot \sigma_{\bar x}

   =225000+(0.84\times 15811.39)\\\\=225000+13281.5676\\\\=238281.5676\\\\\approx 238281.57

Thus, the amount of money separating the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings is $238,281.57.

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