A technician is dispatched to install additional RAM in a computer. The technician unplugs the system from the power source, and
1 answer:
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Answer:
A.) in order traversal - 9 32 35 14 27 25 2 7 29
B.) pre-order traversal - 25 14 32 9 35 27 7 2 29
C.) post-order traversal - 9 35 32 27 14 2 29 25
Explanation:
Given - 25, 14, 7, 32, 27, 2, 29,9,35
To find - On a separate electronic page or sheet of paper:
1. List the values using in order traversal
2. List the values using pre-order traversal
3. List the values using post-order traversal
Proof -
Given values are -
25, 14, 7, 32, 27, 2, 29,9,35
1.)
In order is -
9 32 35 14 27 25 2 7 29
2.)
Pre-order is - (Root left right)
List nodes of first time visit
25 14 32 9 35 27 7 2 29
3.)
Post-order is - (Left Right Root)
List nodes of third time visit
9 35 32 27 14 2 29 25
Answer:
The correct Answer is 0.0571
Explanation:
53% of U.S. households have a PCs.
So, P(Having personal computer) = p = 0.53
Sample size(n) = 250
np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10
So, we can just estimate binomial distribution to normal distribution
Mean of proportion(p) = 0.53
Standard error of proportion(SE) = =
= 0.0316
For x = 120, sample proportion(p) = =
= 0.48
So, likelihood that fewer than 120 have a PC
= P(x < 120)
= P( p^ < 0.48 )
= P(z < ) (z=
)
= P(z < -1.58)
= 0.0571 ( From normal table )