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Arturiano [62]
3 years ago
9

Consider the universal set R,R, define the interval A=[−7,1],A=[−7,1], interval B=(−1,5),B=(−1,5), and CC be the negative real n

umbers. Complete the following exercises in interval notation.
1. A∪B=A∪B=

2. A∩B∩C=A∩B∩C=

3. A∩(B∪C)=A∩(B∪C)=

4. A^c ∪ B^c∪ C^c=

5. (C−A)∪(B−C)=(C−A)∪(B−C)=

Mathematics
1 answer:
AURORKA [14]3 years ago
8 0
Draw a diagram representing the real number line, and the sets A  and B.

According to the diagram:

<span>1. A∪B= [-7, 5)

2. </span><span>2. A∩B∩C= (-1, 0)

</span>3. A∩(B∪C)= [-7, 1] <span>∩ (-INFINITY, 5) = [-7, 1]

4. </span>4. A^c ∪ B^c∪ C^c= (not A) ∪ (not B) ∪ (not C) = 

(-infinity, -7) ∪ (1, infinity)   ∪  (-infinity, -1] ∪ [5, infinity)   ∪ [0, infinity)

= (-infinity, -1] ∪ (1, infinity ) ∪ [0, infinity) = (-infinity, -1] ∪ [0, infinity) 

5. (C−A)∪(B−C)= "in C but not in A" union "in B but not in C"

=(-infinity, -7) ∪ [0, 5) 

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there no data

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Find the median of the first and third quartiles81, 69 ,90, 79, 59, 125, 95, 116, 183, 97, 92, 50, 122, 55, 9250, 55, 59, 69, 79, 81, 90,   92,     92, 95, 97, 116, 122, 125, 183.The first quartile: 50, 55, 59, 69, 79, 81, 90: median of the first quartile is 69The third quartile: 92, 95, 97, 116, 122, 125, 183: median of the third quartile is 116.The interquartile range is the difference between the median of both the first and third quartile:  which is 116 - 69 = 47.There are no outliers in this set of data

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Step-by-step explanation:

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