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NEW QUESTION: 1
-- Exhibit --- Exhibit -
An administrator receives notice of a large number of ARP table issues. The issues began after a maintenance window 36 hours ago. The issue did not exist prior to the maintenance window.
Which change was made to the port group that is causing the ARP table issue?
A. Failback was changed to Yes.
B. Uplink 2 was changed from Standby to Active.
C. Network failure detection was changed to Link status only.
D. Notify Switches was changed to No.
Answer: D
NEW QUESTION: 2
A company uses Dynamics 365 Process manufacturing.
Employees work a half day so that they can attend a company event. You have a rush order that must be completed that same day.
After reviewing the net requirements of the batch order, only 400 of the total quantity of 1400 units must be completed today. The production manager must retain the original batch number as the quantity of 400. The original batch number must appear on documentation for the customer.
You need to split the batch order into two orders.
Which three actions should you perform in sequence? To answer, move the appropriate actions from the list of actions to the answer area and arrange them in the correct order.
Answer:
Explanation:
1 - Select the batch order.
2 - Select Split
3 - Enter 1,000 in the split quantity field.
NEW QUESTION: 3
View the exhibit and examine the structure of the EMPLOYEES table.
You want to display all employees and their managers having 100 as the MANAGER_ID.
You want the output in two columns: the first column would have the LAST_NAME of the managers and the second column would have LAST_NAME of the employees.
Which SQL statement would you execute?
A. 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;
B. 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;
C. 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;
D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id and AND e.manager_id = 100
Answer: C