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11111nata11111 [884]
3 years ago
9

Solve the system of linear equations by graphing.

Mathematics
2 answers:
inn [45]3 years ago
6 0

The solution is where the two lines cross each other which is (2.3, 4.1)

To graph the lines replace x with a number and solve for y, so this at least twice for each equation. This will give you points to use to draw the lines.

sasho [114]3 years ago
4 0

Answer:

2.3/4.1

Step-by-step explanation:

Took it on Edg, it was right.

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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
Number of elements in set E^{c} = n(E^{c})=3

So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ &#10;P(E^{c})= \frac{3}{12} \\  \\ &#10;P(E^{c})= \frac{1}{4}
7 0
3 years ago
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