QUESTION IMAGE
Question
use the given information to fill in the number of elements for each region in the venn diagram.
$n(a) = 35$, $n(b) = 29$, $n(a cup b) = 45$, $n(a cap b) = 11$
the number of elements in region $x$ is $\square$, the number of elements in region $y$ is $\square$, the number of elements in region $z$ is $\square$, and the number of elements in region $w$ is $\square$.
(type whole numbers.)
Step1: Use De - Morgan's law
By De - Morgan's law, \(n(A'\cup B')=n((A\cap B)')\). And we know that \(n((A\cap B)') = N - n(A\cap B)\), where \(N\) is the universal set. So \(N=n((A\cap B)')+n(A\cap B)\). Since \(n(A'\cup B') = 45\) and \(n(A\cap B)=11\), then \(N=45 + 11=56\).
Step2: Find \(n(A)\)
We know that \(n(A')=35\), and by the formula \(n(A')=N - n(A)\), so \(n(A)=N - n(A')\). Substituting \(N = 56\) and \(n(A')=35\), we get \(n(A)=56 - 35=21\).
Step3: Find the elements in each region
- Region \(y\): Given \(n(A\cap B) = 11\), so \(y = 11\).
- Region \(x\): Since \(n(A)=x + y\) and \(n(A)=21,y = 11\), then \(x=n(A)-y=21 - 11 = 10\).
- Region \(z\): Given \(n(B)=29\) and \(n(B)=y + z\), so \(z=n(B)-y\). Substituting \(y = 11\) and \(n(B)=29\), we get \(z=29 - 11=18\).
- Region \(w\): Given \(n(A')=35\) and \(n(A')=w+z\), so \(w=n(A')-z\). Substituting \(n(A') = 35\) and \(z = 18\), we get \(w=35 - 18 = 17\).
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The number of elements in region \(x\) is \(10\), the number of elements in region \(y\) is \(11\), the number of elements in region \(z\) is \(18\), and the number of elements in region \(w\) is \(17\).