Differences between two samples are least likely to be statistically significant if the samples are _______ and the standard dev
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Answer:
Step-by-step explanation:
we have
Equate to zero
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
so
substitute in the formula
Remember that
so
Answer:
(a) Shown below.
(b) The probability that the first ball drawn is blue is 0.40.
(c) The probability that only white balls are drawn is 0.36.
Step-by-step explanation:
The balls in the urn are as follows:
Blue balls: B₁ and B₂
White balls: W₁, W₂ and W₃
It is provided that two balls are drawn from the urn, with replacement, and their color is recorded.
(a)
The possible outcomes of selecting two balls are as follows:
B₁B₁ B₂B₁ W₁B₁ W₂B₁ W₃B₁
B₁B₂ B₂B₂ W₁B₂ W₂B₂ W₃B₂
B₁W₁ B₂W₁ W₁W₁ W₂W₁ W₃W₁
B₁W₂ B₂W₂ W₁W₂ W₂W₂ W₃W₂
B₁W₃ B₂W₃ W₁W₃ W₂W₃ W₃W₃
There are a total of N = 25 possible outcomes.
(b)
The sample space for selecting a blue ball first is:
S = {B₁B₁, B₁B₂, B₁W₁, B₁W₂, B₁W₃, B₂B₁, B₂B₂, B₂W₁, B₂W₂, B₂W₃}
n (S) = 10
Compute the probability that the first ball drawn is blue as follows:
Thus, the probability that the first ball drawn is blue is 0.40.
(c)
The sample space for selecting only white balls is:
X = {W₁W₁, W₂W₁, W₃W₁, W₁W₂, W₂W₂, W₃W₂, W₁W₃, W₂W₃, W₃W₃}
n (X) = 9
Compute the probability that only white balls are drawn as follows:
Thus, the probability that only white balls are drawn is 0.36.
we are given
Firstly, we will split terms
We can group first two and last two terms
we can see that 2x^2 is common in first two terms
and 3 is common in last two terms
so, we can factor factor out 2x^2 from first group and 3 from second group
we get
we can see that each terms are multiple of 6x-1
so, we can factor out 6x-1
so, we get
............Answer