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Len [333]
3 years ago
9

Please help me simplify the expression on number 19!!!

Mathematics
2 answers:
ANTONII [103]3 years ago
6 0

Answer:

Step-by-step explanation:

First you have to do the multiplication

y+4+3(y+2)  Use the distributive property to multiply 3(y+2)  

3 times y equals 3y  and 3 times 2 equals 6, so,

3(y+2)=3y+6  Put it into the equation...

y+4+3y+6          Now add like terms.(I like to gether the like terms together before adding.)  y+3y +4+6  

y is 1y, so y+3y=4y   and 4+6=10

So, y+3y+4+6=4y+10

y+4+3(y+2)=4y+10

777dan777 [17]3 years ago
3 0

We can start off by using the distribution method:

y + 4 + 3(y + 2)

Multiply everything in the parentheses by 3:

y + 4 + 3*y + 3*2

= y + 4 + 3y + 6

Simplify:

4y +  10

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Assume that 12 people, including the husband and wife pair, apply for 4 sales positions. People are hired at random.
Iteru [2.4K]

brainly.com/question/5218999


The formula C(n, r)= \frac{n!}{r!(n-r)!}, where r! is 1*2*3*...r

is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.

for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.

-----------------------------------------------------------------------------------------------


Selecting 4 people out of 12 can be done in :

\displaystyle{C(12, 4)= \frac{12!}{4!8!}= \frac{12\cdot11\cdot10\cdot9\cdot8!}{4!8!}= \frac{12\cdot11\cdot10\cdot9}{4!}=11\cdot5\cdot9= 495       many ways.


All the possible groups of 4 people, where the husband and wife are included, can be done in C(10, 2) many ways, since we only calculate the possible choices of 2 out of 10 people, to complete the groups of 4.


\displaystyle{ C(10, 2)= \frac{10!}{2!8!}= \frac{10\cdot9}{2}=45


Thus, the 

probability that both the husband and wife are hired is 45/495=0.09


Part 2)

The probability that one is hired and the other is not = 

P(husband hired, wife not hired) + P(wife hired, husband not hired)

these 2 are clearly equal, so it is enough to calculate one.


Consider the case : husband hired, wife not hired.

assuming the husband is hired, we have to calculate the possible groups of 3 that can be formed from 11-1 (the wife)=10 people.

this is 

\displaystyle{ C(10, 3)= \frac{10!}{3!7!}= \frac{10 \cdot9 \cdot8}{3\cdot2}=10\cdot3\cdot4=120


thus, 


P(husband hired, wife not hired)=120/495=0.24


thus, 

The probability that one is hired and the other is not = 

P(husband hired, wife not hired) + P(wife hired, husband not hired) =

0.24+0.24=0.48



Answer:


A) 0.09


B) 0.48

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