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Delicious77 [7]

1 year ago

9

Pls answer it fast i will mark you as brainiest.

1 answer:

jeka941 year ago

8 0

$\tt -1\dfrac{3}{4}=-\dfrac{(1\times 4)+3}{4}=-\dfrac{7}{4}$

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Using the following information = Universal set = {2,3,4,5,6,7,8,9,12,13,14,16,20,22,56). Subset A = {9,12,13,20,22,56); Subset

Cloud [144]

$\dfrac{2}{5},\dfrac{13}{15},\dfrac{3}{5},\dfrac{1}{5},\dfrac{1}{5}$

step-by-step explanation:

The intersection of two sets A and B, denoted by A ∩ B, is the set containing all elements of A that also belong to B.

The union (denoted by ∪) of a collection of sets is the set of all elements in the collection.

$P(s)$ of a set $s$ is defined as ratio of number of elements in $s$ to the number of elements in $universal set$

given $Universalset=$ $\{$ $\text{2,3,4,5,6,7,8,9,12,13,14,16,20,22,56}$ $\}$

given $A=\{9,12,13,20,22,56\}$ and $H=\{4,5,8,9,16,22\}$

and $C=\{1,4,20,22,56\}$

For Question A:

$A$ ∩ $H$ = $\{9,12,13,20,22,56\}$ ∩ $\{4,5,8,9,16,22\}$

= $\{9,22}\}$

( $A$ ∩ $H$ )∪ $C$ = $\{9,22}\}$ ∪ $C=\{1,4,20,22,56\}$ = $\{1,4,9,20,22,56\}$

$p((A$ ∩ $H)$ ∪ $C)$ = $\frac{6}{15}=\frac{2}{5}$

For Question B:

$H$ ∩ $C$ = $\{4,22\}$

$p(H$ ∩ $C)$ '= $\frac{15-2}{15}=\frac{13}{15}$

For Question C:

$p(H)$ '= $\frac{15-6}{15}=\frac{3}{5}$

For Question D:

$C$ \ $H$ = $\{1,20,56\}$

$p(C$ \ $H)$ = $\frac{3}{15}=\frac{1}{5}$

For Question E:

$A$ \ $C$ = $\{9,12,13\}$

$p(A$ \ $C)$ = $\frac{3}{15}=\frac{1}{5}$

7 0

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