What is the end behavior of f(x) = -0.5^2 + 2
1 answer:
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Answer:
B, C, F
Step-by-step explanation:
Given:
Subtract the second equation from the first equation:
Option B "The x coordinate of the solution is 9" is ture
Substitute it into the first equation:
Option F "The y coordinate of the solution is -8" is true.
Point (9,-8) is in the IV quadrant, so option C is true.
Answer:
The expected number of experiment is 198
Step-by-step explanation:
Solution
Given that:
We need to carry out a test
where,
H0 : p =0.5
and
H1 : p ≠ 0.5
n = the number of flip coin which is = 200
x = this is the number of heads declared = 106
So,
xₙ = x/n = 106/200 = 0.53 = p
Thus,
D₂ =√n (xₙ - 0.5)/√0.5 * (1-0.5)
=√200 * (0.53 - 0.5)/ √0.5 * (1-0.5)
= 0.848528137
D₂ = 0.8485
Now,
p ( z> D₂ ) = p ( z > 0.8485)
=0.198072
Thus,
By applying R,
1 - pnorm (0.8485, 0,1)
That is (1- pnorm (D₂, 0, 1)
Hence,
p ( z> D₂ )≈ 0.198072
So,
We find The expected number of experiment such that the estimator √n (xₙ - 0.5)/√0.5 * (1-0.5)i s larger than the value D₂ when the total is 1000 times or attained in the first experiment Thus
1000 * p ( z> D₂ )
= 1000 * 0.198072
=198.072
=198
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