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Nat2105 [25]
3 years ago
6

If g(x) = -3x + 12, what is the solution set if the domain of gis (-2, 0,2}?

Mathematics
1 answer:
disa [49]3 years ago
4 0

Given:

The function is

g(x)=-3x+12

Consider the given domain of g is {-2,0,2}.

To find:

The solution set  if the domain of g for the given domain.

Solution:

Domain is the set of input values and range is the set of output value.

We have,

g(x)=-3x+12

Domain of g is {-2,0,2}.

For x=-2,

g(-2)=-3(-2)+12

g(-2)=6+12

g(-2)=18

For x=0,

g(0)=-3(0)+12

g(0)=0+12

g(0)=12

For x=2,

g(2)=-3(2)+12

g(2)=-6+12

g(2)=6

Now, the solution set  of the given domain of g is

\{f(-2),f(0),f(2)\}=\{18,12,6\}

Therefore, the solution set is {18,12,6}.

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1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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3 years ago
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