Write the specified set using the roster method given.U={a,b,c,d,e,f,g,h},M={a,b,d},N={b,d,f,h}U={\left\lbrace a,b,c,d,e,f,g,h\right\rbrace},M={\left\lbrace a,b,d\right\rbrace},N=\left\lbrace{b,d,f,h}\right\rbrace.N′∩(M∪U)N^{\prime}\cap(M\cup U)

N′∩(M∪U)={a,c,e,g}$$N^{\prime}$$\cap(M\cup U)=\left\lbrace a,c,e,g\right\rbrace

Write the specified set using the roster method given. U = { a , b , c , d , e , f , g , h } , M = { a , b , d } , N = { b , d , f , h } U={\left\lbrace a,b,c,d,e,f,g,h\right\rbrace},M={\left\lbrace a,b,d\right\rbrace},N=\left\lbrace{b,d,f,h}\right\rbrace . N ′ ∩ ( M ∪ U ) N^{\prime}\cap(M\cup U)

Welcome back. For this problem, we want to find N complement intersect M union U and write it in roster notation. And to help us out, we've been given our universal set U, as well as sets M and N. Now, we need to start in the parentheses with M union U. And M union U is going to be the set of all elements in M. In you or in both, but you are universal set. It contains everything. Everything is in you. So if everything is in you, then everything is inside of M, U, or both, meaning that everything is inside of our union, M union you. So this is equal to the universal set, which includes the elements A, B, C, D, E, F, G, and H. Now that we have M union U, we want to find the intersection of N complements. And M Union you. But that means first, we have to find N complement. Now, as a reminder, and complement is going to be all of the elements in the universal set, but not in N. For example, A is in the universal set, but not in N, so it will be an N complement, but B is an N, so it's not gonna be an N complement. C is not an N, so it will be an N complement, but D is an N, so it won't be. E is not in N, so it will be an N complement, but F is, so it will not be, and G is not in N, so it will be an N complement, but H is, so it will not be. So we get that N complement is the set containing A, C, E, and G. Now, we want to find the intersection of N complement and M union U. So we just need to check which elements are in N complements that are also in MUion U. Now, we can see that A. See E and G, all of our elements of N complements, are also in MUion U, which makes sense because MUU is the universal set. Since all of our elements in N complement are also in MUU, they're going to be in the intersection of N complement and MUion U. So we can say that this intersection is a set containing the elements A, C, E, and G. We don't need to check any other elements, as any elements in the intersection have to be in and complement. Great work.

Write the specified set using the roster method given.U={a,b,c,d,e,f,g,h},M={a,b,d},N={b,d,f,h}U={\left\lbrace a,b,c,d,e,f,g,h\right\rbrace},M={\left\lbrace a,b,d\right\rbrace},N=\left\lbrace{b,d,f,h}\right\rbrace.N′∩(M∪$$U)N^{\prime}$$\cap(M\cup U)

N′∩(M∪U)={a,c,e,g}$$N^{\prime}$$\cap(M\cup U)=\left\lbrace a,c,e,g\right\rbrace

N′∩(M∪U)={b,d,f,h}$$N^{\prime}$$\cap(M\cup U)=\left\lbrace b,d,f,h\right\rbrace

N′∩(M∪U)={a,b,d}$$N^{\prime}$$\cap(M\cup U)=\left\lbrace a,b,d\right\rbrace

N′∩(M∪U)={a,b,c,d,e,f,g,h}$$N^{\prime}$$\cap(M\cup U)=\left\lbrace a,b,c,d,e,f,g,h\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.
$U = {a, b, c, d, e, f, g, h}, M = {a, b, d}, N = {b, d, f, h}$ .
$N^{'} \cap (M \cup U)$

$N^{'} \cap (M \cup U) = {a, c, e, g}$

$N^{'} \cap (M \cup U) = {b, d, f, h}$

$N^{'} \cap (M \cup U) = {a, b, d}$

$N^{'} \cap (M \cup U) = {a, b, c, d, e, f, g, h}$

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Struggling with Quantitative Reasoning?

Write the specified set using the roster method given.U={a,b,c,d,e,f,g,h},M={a,b,d},N={b,d,f,h}U={\(\left\]\lbrace\) a,b,c,d,e,f,g,h\(\right\[\rbrace\)},M={\(\left\]\lbrace\) a,b,d\(\right\[\rbrace\)},N=\(\left\]\lbrace{b,d,f,h}\[\right\]\rbrace\).N′∩(M∪U)N^{\(\prime\)}\(\cap\)(M\(\cup\) U)

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Write the specified set using the roster method given.
$U = {a, b, c, d, e, f, g, h}, M = {a, b, d}, N = {b, d, f, h}$ .
$N^{'} \cap (M \cup U)$