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)=
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)=
1 answer:
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)
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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Answer:
1- A critical value right-tailed Normality test.
2- Yes, they did.
3-
<u>Null hypothesis </u>
The national average on the statistics achievement test is 60
<u>Alternative hypothesis </u>
The average on the experimental program for statistics education is greater than 60
4- 0.05 or 5%
5- A Type 1 Error is the error we make when we reject the null hypothesis given that it is true.
Step-by-step explanation:
1)
Since the sample size is greater than 30, the population follows a Normal distribution, and we suspect the average is greater than the established one, the appropriate test to use is a critical value right-tailed Normality test.
2)
To check if the students scored significantly above the national average in this special program to a significance level α = 0.05, we must check if the <em>z-statistic</em> given by the sample falls to the right of 1.64, since the area under the Normal N(0;1) to the right of 1.64 equals 0.05
<em>Our z-statistic is given by the formula </em>
<em>
</em>
<em>where </em>
<em> is the mean of the sample
</em>
<em> is the mean of the null hypothesis
</em>
<em> is the standard deviation
</em>
<em>n is the sample size </em>
Our z-statistic is then
<em>Since 2.4 > 1.64, the students did score significantly above the national average in this special program to a significance level α = 0.05 </em>
3)
<u>Null hypothesis </u>
The national average on the statistics achievement test is 60
<u>Alternative hypothesis </u>
The average on the experimental program for statistics education is greater than 60
4)
The probability of a Type 1 Error is 0.05 or 5% (the significance level)
5)
A Type 1 Error is the error we make when we reject the null hypothesis given that it is true.