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Anestetic [448]
3 years ago
8

Can someone please help me.......

Mathematics
1 answer:
svetoff [14.1K]3 years ago
5 0
Point -2,-1 would change to -2,0. the area of the new rectangle is 10 square units
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Does 3/4 have the same value as the expression 4 ÷ 3 answer
Fed [463]
No. 4/3 or 4 divided by three is not the same as 3/4 or three fourths. 4/3 is 1.33333333333. 3/4 is .75
3 0
3 years ago
What are the domain and range of the function represented by the table?
andrew-mc [135]

Answer:

A. Domain: {-8, -6, -4, -2, 0} Range: {2}

Step-by-step explanation:

The domain of a function is the x values, the range is the y values. you order them in either greatest to least or least to greatest (keep this consistent in both domain and range values). also, if a value repeats, only list it once (this is why the range is {2} rather than {2, 2, 2, 2, 2} )

5 0
3 years ago
8
givi [52]

Answer:

8

Step-by-step explanation:

8 is the digit after 7  and previous to 9

8 0
3 years ago
I NEED HELP ASAP PLZ
FinnZ [79.3K]

first one

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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

4 0
3 years ago
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