Write the specified set using the roster method givenU={1,2,3,…,20},E={2,4,6,…,20},F={5,10,15,20}U={\left\lbrace1,2,3,\ldots,20\right\rbrace},E={\left\lbrace2,4,6,\ldots,20\right\rbrace},F={\left\lbrace5,10,15,20\right\rbrace}E′∪FE^{\prime}\cup F

E′∪F={1,3,5,7,9,10,11,13,15,17,19,20}$$E^{\prime}$$\cup F=\left\lbrace1,3,5,7,9,10,11,13,15,17,19,20\right\rbrace

Write the specified set using the roster method given U = { 1 , 2 , 3 , … , 20 } , E = { 2 , 4 , 6 , … , 20 } , F = { 5 , 10 , 15 , 20 } U={\left\lbrace1,2,3,\ldots,20\right\rbrace},E={\left\lbrace2,4,6,\ldots,20\right\rbrace},F={\left\lbrace5,10,15,20\right\rbrace} E ′ ∪ F E^{\prime}\cup F

Welcome back. For this problem, we've been given our universal set U, as well as sets E and F, and we want to find E complements union F and write it in roster notation. Now, E complements union F is just going to be the set of things either in E complements, in F, or in both. So it's going to be helpful to have the elements of E complements and the elements of F written out, which means our first step here is going to be to find E complements. Now, we can see that U is the set 123, and so on and so forth up to 20. So here we have all of the whole numbers from 1 to 20 in our universal set. If we look in E, we can see that it's the set containing 246, and so on and so forth up to 20. So following this pattern, we can see that it is the set of even numbers from 1 to 20. Meaning that the complement, all of the stuff in the universal set, but not in E, all of the whole numbers from 1 to 20, but not the even ones, is going to be the odd numbers from 1 to 20. So we can say that E complement is going to be the set including 1. 3579 1113, 1517 and 19, all of the odd numbers from 1 to 20, all of the things in the universal set, not in e. Now we do have F written out, so now that we have both of our sets, let's find the union. We're gonna start by adding in the elements from E complements. So, we have that our union is going to include 13579, 1113, 15, and 17 and 19, all of the elements from E complements. Now, we just need to add in the elements from F that we haven't already added into our union. For example, we don't need to add in 5 again, because we've already added it into the union when we were adding the elements in E complement. But because 10 is not already in our union, we do need to add it in. Next we have 15, which we already have in the union, so we don't need to add it again. And finally, we have 20, which is not already in our union, so we do need to add in 20. Now that we've made sure that every element that does appear in E complements, or in F or in both is inside of our union, we're all set, and we can close out our curly braces. We get that the union of E complements and F is a set containing 13579, 1113, 1517, 1910, and 20. Now, you could reorder these elements, but the order of the elements inside of her curly braces doesn't matter, so this is a perfectly acceptable, correct answer. Great work.

Write the specified set using the roster method givenU={1,2,3,…,20},E={2,4,6,…,20},F={5,10,15,20}U={\left\lbrace1,2,3,\ldots,20\right\rbrace},E={\left\lbrace2,4,6,\ldots,20\right\rbrace},F={\left\lbrace5,10,15,20\right\rbrace}E′∪$$FE^{\prime}$$\cup F

E′∪F={1,3,5,7,9,10,11,13,15,17,19,20}$$E^{\prime}$$\cup F=\left\lbrace1,3,5,7,9,10,11,13,15,17,19,20\right\rbrace

E′∪F={1,3,5,7,9,11,13,15,17,19}$$E^{\prime}$$\cup F=\left\lbrace1,3,5,7,9,11,13,15,17,19\right\rbrace

E′∪F={5,10,15,20}$$E^{\prime}$$\cup F=\left\lbrace5,10,15,20\right\rbrace

E′∪F={2,4,6,8,10,12,14,16,18,20}$$E^{\prime}$$\cup F=\left\lbrace2,4,6,8,10,12,14,16,18,20\right\rbrace

2. Sets

Intersections and Unions

Quantitative Reasoning

2. Sets

Intersections and Unions

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Multiple Choice

Write the specified set using the roster method given
$U = {1, 2, 3, \dots, 20}, E = {2, 4, 6, \dots, 20}, F = {5, 10, 15, 20}$
$E^{'} \cup F$

$E^{'} \cup F = {1, 3, 5, 7, 9, 10, 11, 13, 15, 17, 19, 20}$

$E^{'} \cup F = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}$

$E^{'} \cup F = {5, 10, 15, 20}$

$E^{'} \cup F = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}$

3:51m

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