/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 The rise time of a reactor is me... [FREE SOLUTION] | 91Ó°ÊÓ

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The rise time of a reactor is measured in minutes (and fractions of minutes). Let the sample space be positive, real numbers. Define the events \(A\) and \(B\) as follows: \(A=\\{x \mid x<72.5\\}\) and \(B=\\{x \mid x>52.5\\}\) Describe each of the following events: (a) \(A^{\prime}\) (b) \(B^{\prime}\) (c) \(A \cap B\) (d) \(A \cup B\)

Short Answer

Expert verified
(a) \( A' = \{x \mid x \geq 72.5\} \); (b) \( B' = \{x \mid x \leq 52.5\} \); (c) \( A \cap B = \{x \mid 52.5 < x < 72.5\} \); (d) \( A \cup B = \mathbb{R}^+ \).

Step by step solution

01

Understand Complement of Set A

To find the complement of event A, denoted as \( A' \), we need to determine the set of all elements that are not in A. Given that \( A = \{x \mid x < 72.5\} \), the complement \( A' \) includes all elements greater than or equal to 72.5. Thus, \(A' = \{x \mid x \geq 72.5\}\).
02

Determine Complement of Set B

For the complement of event B, denoted as \( B' \), we consider all elements that are not part of B. Since \( B = \{x \mid x > 52.5\} \), the complement \( B' \) includes all elements less than or equal to 52.5. Therefore, \(B' = \{x \mid x \leq 52.5\}\).
03

Find Intersection of Sets A and B

The intersection \( A \cap B \) represents elements common to both A and B. Event A includes numbers less than 72.5, and event B includes numbers greater than 52.5. Thus, the intersection \( A \cap B = \{x \mid 52.5 < x < 72.5\}\).
04

Determine Union of Sets A and B

For the union \( A \cup B \), we combine all elements from both sets A and B. This set includes numbers satisfying either \( x < 72.5 \) or \( x > 52.5 \). Consequently, the union \( A \cup B = \{x \mid x < 72.5 \text{ or } x > 52.5\}\), which covers all positive real numbers except precisely 72.5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complement of a Set
In set theory, the complement of a set includes all the elements that are not part of the original set. It's like finding the opposite or what's missing in a set from the universal set. Using this for event sets, like in our example for event A, where event A includes all elements less than 72.5:
  • The complement of A, denoted as \( A' \), are elements greater than or equal to 72.5.
Similarly for event B, which includes items greater than 52.5:
  • The complement of B, \( B' \), includes all items less than or equal to 52.5.
In simpler terms, if you know what is inside a set, the complement contains everything else not in that set. This concept is particularly useful when determining set differences.
Intersection of Sets
The intersection of two sets consists of the elements that belong to both sets at the same time. It’s like finding common ground between two groups. For events A and B:
  • Event A includes numbers below 72.5.
  • Event B includes numbers above 52.5.
The intersection \( A \cap B \) will include the numbers that are greater than 52.5 but less than 72.5. This is because these are the values that fit into both conditions set by A and B at the same time.Breaking it down makes it clear: intersection focuses on shared members between sets, narrowing down to specific, mutual components.
Union of Sets
The union of sets brings together all elements from both sets, similar to combining everything into one bigger set. With union, an element is included if it is in either or both sets. Looking at our events A and B:
  • Event A holds numbers less than 72.5.
  • Event B holds numbers greater than 52.5.
The union \( A \cup B \) is the set of numbers where either condition is met: those less than 72.5 or more than 52.5. Essentially, it covers everything except numbers precisely equal to 72.5. This expansive approach makes it straightforward to gather and combine all possible outcomes from both sets, painting a complete picture.

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