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Exam Code:
MLA-C01
Exam Name:
AWS Certified Machine Learning Engineer - Associate
Questions:
94 Q&A
Product Type:

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NEW QUESTION: 1
You are configuring a drop-down list with a page list property. When the user selects an item in the list, you
want to retrieve data associated with the selected item and add the data to the case.
How do you configure the data access for the page list property?
A. Select the Refer to a data page option so the selected information can be retrieved using a data page
lookup.
B. Select the Manual option so the user can add selected data to the page-type property through the UI.
C. Select the Copy data from a data page option so the selected information can be retrieved using a data
page lookup.
D. Select the Allow use as reference property in activities option so the selected information can be
retrieved using an activity.
Answer: B

NEW QUESTION: 2

<input name = 'credentials' type = 'TEXT' value = '"+

A. Option D
B. Option A
C. Option B
D. Option C
E. Option E
Answer: B
Explanation:


NEW QUESTION: 3
Which two principles must be considered when deploying a wireless network through a wooded area? (Choose two.)
A. refraction
B. diffraction
C. absorption
D. reflection
E. scattering
Answer: C,E

NEW QUESTION: 4
You have two lists of values to correlate.

Which query lists all names in colors1 and how many total matches are there in colors2?
A. SELECT colors1.name, count (colors2.name)
FROM JOIN colors2
on colors1.name = colors2.name
GROUP BY colors1.name;
B. SELECT colors1.name, count (colors2.name)
FROM colors1
RIGHT JOIN colors1
on colors1.name =colors2.name
GROUP BY colors1.name;
C. SELECT colors1.name, count(colorse2.name)
FROM colors1.name = colors2.name
WHERE colors1.name =colors2.name
GROUP BY colors1.name;
D. SELECT colors1.name.count (colors2.name)
FROM colors1.colors2
WHERE colors1.name = (SELECT DISTINCT name FROM colors2
WHERE colors1.name=colors2.name)
GROUP BY colors1.name;
E. SELECT colors1.name count (colors2.name)
FROM colors1
INNER JOIN colors2
on colors1.name =colors2.name
GROUP BY colors1.name;
Answer: A

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