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NEW QUESTION: 1
You are responsible for maintaining and troubleshooting PC's at your company. The receptionist reports her screen has gone blue. When you get there you notice the 'blue screen of death' with an error message NTFS_FILE_SYSTEM. What is the most likely cause of this error?
A. Get the latest patch for Windows.
B. Windows was installed improperly.
C. A virus
D. The hard disk is corrupt
Answer: D

NEW QUESTION: 2
회사에 기존 온 프레미스 3 계층 웹 애플리케이션이 있습니다. Linux 웹 서버는 콘텐츠가 다양한 소스에서 하루에 여러 번 새로 고쳐지기 때문에 NAS 서버의 중앙 집중식 파일 공유에서 콘텐츠를 제공합니다. 기존 인프라는 최적화 되지 않았으며 회사는 부하에 대응하여 리소스를 확장 및 축소 할 수있는 기능을 얻기 위해 AWS로 이동하려고 합니다. 온 프레미스 및 AWS 리소스는 AWS Direct Connect를 사용하여 연결됩니다.
회사가 콘텐츠 새로 고침 프로세스를 지연하지 않고 웹 인프라를 AWS로 마이그레이션하려면 어떻게 해야 합니까?
A. Auto Scaling 그룹의 AWS에서 웹 서버 Amazon EC2 인스턴스를 생성합니다. 웹 서버 인스턴스가 NAS 서버에서 업데이트되는 야간 프로세스를 구성합니다.
B. AWS Storage Gateway를 사용하여 온 프레미스 파일 게이트웨이를 생성하여 NAS 서버를 교체하고 콘텐츠를 AWS에 복제합니다. AWS 측에서는 동일한 Storage Gateway 버킷을 각 웹 서버 Amazon EC2 인스턴스에 탑재하여 콘텐츠를 제공합니다.
C. AWS의 Classic Load Balancer 뒤에 웹 서버 Amazon EC2 인스턴스 클러스터를 생성합니다. 콘텐츠에 대한 모든 인스턴스간에 Amazon EBS 볼륨을 공유합니다. 이 볼륨과 NAS 서버의 주기적인 동기화를 예약합니다.
D. NAS 서비스 역할을 할 Amazon EFS 공유를 온 프레미스 사용자에게 노출합니다. 동일한 EFS 공유를 웹 서버 Amazon EC2 인스턴스에 탑재하여 콘텐츠를 제공합니다.
Answer: D
Explanation:
File gateway is limited by performance its gateway instance, whether EC2 or On-premises, Cache will get filled up fast if not properly configured, For large number of EC2 instances EFS scales better. So, bottom line is File Storage gateway is for legacy applications and you have to add cost of large gateway instances before comparing it to same quantity of EFS storage. https://www.reddit.com/r/aws/comments/82pyop/storage_gateway_vs_efs/
https://docs.aws.amazon.com/efs/latest/ug/efs-onpremises.html

NEW QUESTION: 3
A developer is writing an application with three java Persistence API entities: order, customer, and Address. There is a many-to-one relationship between order and customer, and a one to-many relationship between customer and Address.
Which two Criteria queries will return the orders of all customers who have an address whose value specified by the String parameter postalcode? (Choose two)
A. String postalCode = . . .
Criteria Builder cb = . . .
Root <Order> order = cq.from (Order.class);
order.join (order_. customer).join(Customer_.addresses);
cq.where (cb.equal (address.get(Address_.postalCode), postalCode));
cq.select (order). Distinct (true);
// query execution code here
B. String postalCode = ...
CritetiaBuilder cb = ...
Root<order> order = cq - from (Order . class) ,
Join<order, Address> address = order.join(Customer_.addresses);
cq.where(ct>.equal(address.get(Address_.postalCode), postalCode));
cq-select(order).distinct(true);
// query execution code here
. . .
C. String postalCode = ...
CriteriaBuilder cb = ...
Root<order> order = cq- from (Order . class ) ,
Join<order, Address> address = order . join (Order_. customer) -join (Customer_.addresses);
cq.where <cb.equal (address.get(Address_.postalCode) , postalCode) ) ;
cq.selec:(order).distinct(true);
// query execution code here
. . .
D. String postalCode = . . .
Criteria Builder cb = . . .
CriteriaQuery<order> cq = cb.createQuery (Order.class);
Root <order> order = cq.from(order.class);
Join <order, Customer> customer = order.join(Order_.customer);
Root <Order> order = cq.from (Order.class);
Join <customer, Address> address = customer join (Order_.customer)
cq.where (cb.equal (address.get(Address_.postalCode), postalCode));
cq.select (order). Distinct (true);
// query execution code here
. . .
Answer: B,C