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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. Cumulative share
B. Ordinary share
C. Preference share
D. Deferred share
Answer: C

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. 500 A
B. 250 A
C. 320 A
D. 630 A
Answer: C

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