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NEW QUESTION: 1
What are two places the domain properties can be found? (Choose two.)
A. TIBCO_HOME/designer/domain
B. TIBCO_HOME/repository
C. TIBCO_HOME/domain
D. TIBCO_HOME/tra/domain
E. TIBCO_HOME/administrator/domain
Answer: D,E

NEW QUESTION: 2
You have a privately addressed email server behind your Firebox. If you want to make sure that all traffic from this server to the Internet appears to come from the public IP address 203.0.113.25, regardless of policies, which from of NAT would you use? (Select one.)
A. Create a global dynamic NAT rule for traffic from the email server and set the source IP address to
203.0.113.25.
B. In the SMTP policy that handles traffic from the email server, select the option to apply dynamic NAT to all traffic in the policy and set the source IP address 203.0.113.25.
C. Create a static NAT action for traffic to the email server, and set the source IP address to 203.0.113.25.
Answer: A

NEW QUESTION: 3
After you provision a new H.323 gateway to integrate with a Cisco Unified Communications Manager cluster, you see that the H.323 gateway shows unregistered in Cisco Unified CM.
Which action is needed to register the H.323 gateway with Cisco Unified Communications Manager?
A. H.323 gateways never register with Cisco Unified Communications Manager. No action is required to address the unregistered state of the H.323 gateway with Cisco Unified CM.
B. Create a VoIP dial peer on the H.323 gateway to point to the Cisco Unified Communications Manager cluster.
C. Stop and restart the Cisco Call Manager service.
D. H.323 gateways must use a loopback address to register to a Cisco Unified Communications Manager cluster.
Answer: A

NEW QUESTION: 4

public class Emp
private String eName;
private Integer eAge;
Emp(String eN, Integer eA) {
this.eName = eN;
this.eAge = eA;
}
public Integer getEAge () {return eAge;}
public String getEName () {return eName;}
}
and code fragment:
List<Emp>li = Arrays.asList(new Emp("Sam", 20), New Emp("John", 60), New Emp("Jim",
51));
Predicate<Emp> agVal = s -> s.getEAge() > 50;//line n1
li = li.stream().filter(agVal).collect(Collectors.toList());
Stream<String> names = li.stream()map.(Emp::getEName);//line n2
names.forEach(n -> System.out.print(n + " "));

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