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NEW QUESTION: 1
展示を参照してください。

ネットワーク管理者は、ネットワーク内のルーターをリモート管理するためにSSHアクセスを許可する必要があります。運用チームは10.20.1.0/25ネットワーク上にあります。どのコマンドがこのタスクを実行しますか？
A. access-list 2699 permit udp 10.20.1.0 0.0.0.255
B. アクセスリストなし2699はtcpを拒否します10.20.1.0 0.0.0.127 eq 22
C. access-list 2699 permit tcp any 10.20.1.0 0.0.0.255 eq 22
D. アクセスリストなし2699 deny ip any 10.20.1.0 0.0.0.255
Answer: D
Explanation:
Note : Already a statement is there in last to allow SSH Traffic for network 10.20.1.0 0.0.0.127, but Second statement says deny ip any 10.20.1.0 0.0.0.255, so how it will work once it is denied. So the right answer is remove the --- no access-list 2699 deny ip any 10.20.1.0 0.0.0.255.

NEW QUESTION: 2
Which of the following statements is false?
A. The external routing information carries a Tag label for transmitting additional information of the route. It is commonly used in a routing policy. The default value of the Tag label is 0.
B. An ABR maintains an LSDB for each area. There are many LSDBs on an ABR.
C. The Link State ID in an ASBR-Summary-LSA is the Router ID of the ASBR.
D. The metric of an ASE LSA can be specified when an external route is imported. The default value is 1.
Answer: A

NEW QUESTION: 3
What BEST describes the value of service design to the business?
A. It provides quick and effective access to standard services
B. It supports the creation of a portfolio of quantified services
C. It reduces total cost of ownership (TCO) of services
D. It improves the control of service assets and configurations
Answer: C

NEW QUESTION: 4
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 e
WHERE m.employee_id = e.manager_id AND e.managerjd=100;
B. SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE e.managerjd=100;
C. SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE m.manager_id=100;
D. SELECT m.last_name "Manager", e.last_name "Employee"
FROM employees m JOIN employees e
ON e.employee_id = m.manager_id
WHERE m.manager_id=100;
Answer: B