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NEW QUESTION: 1
A company issues a share that has a par value of EUR100.00 and pays a fixed dividend of
2.0% annually? What type of share is this?
A. Deferred share
B. Preference share
C. Ordinary share
D. Cumulative share
Answer: B

NEW QUESTION: 2
For two UPS5000-E (50 kVA power module) 150 kVA with 1+1 rack redundancy, a () general input switch is required for the general input power distribution cabinet.
A. 320 A
B. 250 A
C. 630 A
D. 500 A
Answer: A

NEW QUESTION: 3
Which three techniques can be used to improve fault isolation in an enterprise network design? (Choose three.)
A. fully meshed distribution layer
B. tuned Spanning Tree Protocol timers
C. multiple IS-IS flooding domains
D. Equal-Cost Multipath routing
E. aggregate routing information on an OSPF ABR
F. EIGRP query boundaries
Answer: C,E,F
Explanation:
Explanation/Reference:
Explanation:
13

NEW QUESTION: 4
View the exhibit and examine the structure of the EMPLOYEES table.

You want to select all employees having 100 as their MANAGER_ID manages and their manager.
You want the output in two columns: the first column should have the employee's manager's LAST_NAME and the second column should have the employee's LAST_NAME.
Which SQL statement would you execute?
A. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id=100;
B. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id=100;
C. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id AND e.manager_id=100
D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id=100;
Answer: D