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3 years ago

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ujalakhan01 can you answer these? 1. In Spencer’s garden, the number of rose bushes is 7 less than 1.5 times the number of carna

tion bushes. If the number of carnation bushes is c, then the expression representing the number of rose bushes is 2. Using the Distributive Property, the expression 9.78(4.3x + 5.15) simplifies to

1 answer:

aniked [119]3 years ago

7 0

Answer:

B

42.05x + 50.36

Step-by-step explanation:

1. It is 1.5c - 7 because tje mumber of rose bushes is 7 less that 1.5(c), not 1.5(7). So, it is (b).

2. 9.78(4.3x + 5.15)

42.05x + 50.36

Hope this helps:)

P.S. I am not ujalakhan01:(

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

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:

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

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

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