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velikii [3]
2 years ago
6

Select the order in which the vertices are visited in a pre-order traversal of the tree shown below.Question 5 options:e, i, b,

h, f, a, c, d, ge, i, b, h, f, a, d, c, gb, i, f, h, a, d, g, c, eb, i, f, h, d, g, c, a, ePrevious PageNext Page
SAT
1 answer:
Aleksandr [31]2 years ago
3 0

Based on the vertices and the pre-order traversal order the tree would be visited is e, i, b, h, f, a, c, d, g.

<h3>What order will the tree be visited?</h3>

When using the pre-order traversal, the order would be from the root child node to the left child node to right child node.

The root node is the highest node which is e. From here you'll then proceed to the left roots first which means you'll go to i and then b.

From there you go to h and then f. The next nodes would be a and c.

Continuing from left to right, you'll then go to d and then g.

You'll therefore start from "e" and move to "i" and then "b" to "h" and then "f" and "a" to "c" and then to "" and finally to "g"

Find out more on the pre-order traversal at brainly.com/question/14559964.

#SPJ1

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2 years ago
The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and
Mazyrski [523]

Explanation:

Using the normal distribution, it is found that the probability that the volunteer selected will receive a certificate of merit given that the  number of hours the volunteer worked is less than 90 is closest to:

B 0.123

In a normal distribution with mean \muμ and standard deviation \sigmaσ , the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}Z=σX−μ

It measures how many standard deviations the measure is from the mean.  

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of 80, hence \mu = 80μ=80 .

The standard deviation is of 7, hence \sigma = 7σ=7 .

The minimum value is the 80th percentile, which means that it is X_mXm when Z has a p-value of 0.8.

The probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90 is P(X_m < X < 90)P(Xm<X<90) , which is the p-value of Z when X = 90 subtracted by the p-value of Z when X = X_mX=Xm , hence the p-value of Z when X = 90 subtracted by 0.8.

Z = \frac{X - \mu}{\sigma}Z=σX−μ

Z = \frac{90 - 80}{7}Z=790−80

Z = 1.43Z=1.43

Z = 1.43Z=1.43 has a p-value of 0.9236.

0.9236 - 0.8 = 0.1236, hence closest to 0.123, option B.

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