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gayaneshka [121]
3 years ago
8

How do I solve this? Please show work, thank you.

Mathematics
1 answer:
irina [24]3 years ago
7 0

Answer:

a   x sqrt(7) - 49sqrt(x)

Step-by-step explanation:

sqrt(7x) [ sqrt(x) - 7sqrt(7)]

Distribute

sqrt(7x) *sqrt(x) - sqrt(7x)*7sqrt(7)

We know that sqrt(a) sqrt(b)= sqrt(ab)

sqrt(7x^2)  - 7sqrt(7^2 *x)

Now lets separate out the perfect squares

sqrt(7) *sqrt(x^2) - 7sqrt(7^2)*sqrt(x)

x sqrt(7) - 7*7sqrt(x)

x sqrt(7) - 49sqrt(x)

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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 \\  \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\  \\ 
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)} \\  \\ 
P(E^{c})= \frac{3}{12} \\  \\ 
P(E^{c})= \frac{1}{4}
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G(x) = 0 can never happen.

When x is 0, the answer is 4
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When x is 4, the answer is 66.4

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