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sleet_krkn [62]
3 years ago
14

What is the solution to f(x)=g(x) ? Select each correct answer.

Mathematics
2 answers:
lozanna [386]3 years ago
8 0

Answer:

(0,1)

Step-by-step explanation:

To find when f(x)=g(x), then look for a value that is the same for both functions in the table. 1 occurs twice for the same x value. This is when they are equal. The solution is (0,1).

Rzqust [24]3 years ago
5 0

Answer:

The solution of f(x)=g(x) is x=0

Step-by-step explanation:

We need to find the solution of f(x)=g(X) using table.

First we draw the table for difference value of x

 x       f(x)=0.5^x             g(x)=-0.5x+1

 -2       0.5^{-2}=4             -0.5(-2)+1=2

 -1       0.5^{-1}=2             -0.5(-1)+1=\dfrac{3}{2}

 0         0.5^{0}=1               -0.5(0)+1=1

 1          0.5^{1}=\dfrac{1}{2}             -0.5(1)+1=\dfrac{1}{2}

 2         0.5^{2}=\dfrac{1}{4}               -0.5(2)+1=0

Solution of the equation when both function will get same value for same value of x.

0.5^x=-0.5x+1

0.5^0=-0.5(0)+1

1=1

TRUE

In table we can see at x=0

f(0)=1

g(0)=1

f(0)=g(0)=1

Hence, The solution of f(x)=g(x) is x=0

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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 \\  \\ 
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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

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

So, 

P(E^{c})= \frac{n(E^{c}) }{n(S)} \\  \\ 
P(E^{c})= \frac{3}{12} \\  \\ 
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