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OlgaM077 [116]
3 years ago
5

Jose wanted to compare the people in his math class who like country music to those who like rap music. If the Venn diagram abov

e represents the relationship between students who like rap music or country music in Jose's class, which of the following is the best description of regions A, B, and C?
A. A is students who like country music only, B is students who like both country and rap music, and C is students who like rap music only.
B. A all students who like country music, B is all students who like rap music, and C is students who like all types of music.
C. A is students who do not like country music, B is students who like rap music,and C is students who like both country and rap music.
D. A is students who do not like country music, B is students who do not like rap music, and C is students who like country but not rap.
Mathematics
2 answers:
Anestetic [448]3 years ago
6 0
The correct answer to this question is A; <span>A is students who like country music only, B is students who like both country and rap music, and C is students who like rap music only</span>
Roman55 [17]3 years ago
3 0

Answer:  I think it is A personally because that is the only one that sound right to me.

Step-by-step explanation:

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ozzi

Answer: 92 and 94

Step-by-step explanation:

x = first number

x + 2 = second number

x + x + 2 = 186

simplify

2x + 2 = 186

subtract 2 from both sides

2x = 184

divide both sides by 2

x = 92

92 + 2 = 94

The numbers are 92 and 94.

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Step-by-step explanation:

If X is a finite Hausdorff space then every two points of X can be separated by open neighborhoods. Say the points of X are x_1, x_2, ..., x_n. So there are disjoint open neighborhoods U_{12} and U_2, of x_1 and x_2 respectively (that's the definition of Hausdorff space). There are also open disjoint neighborhoods U_{13} and U_3 of x_1 and x_3 respectively, and disjoint open neighborhoods U_{14} and U_4 of x_1 and x_4, and so on, all the way to disjoint open neighborhoods U_{1n}, and U_n of x_1 and x_n respectively. So U=U_2 \cup U_3 \cup ... \cup U_n has every element of X in it, except for x_1. Since U is union of open sets, it is open, and so U^c, which is the singleton \{ x_1\}, is closed. Therefore every singleton is closed.

Now, remember finite union of closed sets is closed, so \{ x_2\} \cup \{ x_3\} \cup ... \cup \{ x_n\} is closed, and so its complemented, which is \{ x_1\} is open. Therefore every singleton is also open.

That means any two points of X belong to different connected components (since we can express X as the union of the open sets \{ x_1\} \cup \{ x_2,...,x_n\}, so that x_1 is in a different connected component than x_2,...,x_n, and same could be done with any x_i), and so each point is in its own connected component. And so the space is totally disconnected.

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