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HCSA-Development–HarmonyIndustry Device V1.0

  • Exam Number/Code : H31-661_V1.0
  • Exam Name : HCSA-Development–HarmonyIndustry Device V1.0
  • Questions and Answers : 213 Q&As
  • Update Time: 2019-01-10
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NEW QUESTION: 1

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

NEW QUESTION: 2
Which of the following is NOT true for risk governance?
A. Risk governance requires reporting once a year.
B. Risk governance is a systemic approach to decision making processes associated to natural and technological risks.
C. Risk governance is based on the principles of cooperation, participation, mitigation and sustainability, and is adopted to achieve more effective risk management.
D. Risk governance seeks to reduce risk exposure and vulnerability by filling gaps in risk policy.
Answer: A
Explanation:
Explanation/Reference:
Explanation:
Risk governance is a continuous life cycle that requires regular reporting and ongoing review, not once a year.
Incorrect Answers:
A, C, D: These are true for risk governance.

NEW QUESTION: 3
View the exhibit and examine the structure in ORDERS and ORDER_ITEMS tables.

You need to create a view that displays the ORDER_ID, ORDER_DATE, and the total number of items in each order.
Which CREATE VIEW statement would create the view successfully?
A. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)||'NO OF ITEMS'FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_dateWHITH CHECK OPTION;
B. CREATE OR REPLACE VIEW ord_vu (order_id, order_date)AS SELECT o.order_id, o.order_date, COUNT(i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
C. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_date;
D. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date,
COUNT(i.line_item_id)FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id,o.order_date;
Answer: C