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polet [3.4K]
2 years ago
14

F(4)= V If g(x) = 2, x = V

Mathematics
1 answer:
gizmo_the_mogwai [7]2 years ago
7 0

By looking at the given graph, we can see that:

  • f(4) = -10
  • g(x) = 2 if x = 0.

<h3>How to identify the value of the function?</h3>

First, we want to find the value f(4). To get that, we need to look at the graph and see the y-value of the function f(x) when x = -4.

By looking at the graph, we can see that f(4) = -10.

Now we know that g(x) = 2, and we want to find the value of x.

Then we need to see the x-value such that g(x) = 2, by looking at the graph of g(x), we can see that it is equal to 2 in the vertex, which is at x = 0.

Then g(x) = 2 when x = 0.

If you want to learn more about graphs:

brainly.com/question/4025726

#SPJ1

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A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c
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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