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yan [13]
3 years ago
14

PLEASE HELP IMMEDIATELY Solve. 3 + q + 9 = 12

Mathematics
2 answers:
Marizza181 [45]3 years ago
8 0
Okay. Combine like terms. You should get 12 + q = 12. subtract 12 from both sides. You'll get 0. q = 0.
vazorg [7]3 years ago
7 0
Add 3+9 and that would be 12
subtract 12-12 and it would be 0
q=0

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Step-by-step explanation:

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olchik [2.2K]
The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in E^{c}. Find P(E^{c}).

Solution:

Part 1:

E^{c} means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

E^{c}=S-E \\  \\ &#10;E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\  \\ &#10;E^{c}=(4,5,6)

Thus the set compliment of E will contain the elements {4,5,6}.So

E^{c} = {4,5,6}

Part 2)

P(E^{c}) means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

P(E^{c}) = (Number of Elements in E^{c})/Number of elements in S

Number of elements in set S = n(S) = 12
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So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ &#10;P(E^{c})= \frac{3}{12} \\  \\ &#10;P(E^{c})= \frac{1}{4}
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