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NEW QUESTION: 1
You need to identify the Volume Licensing program that supports the company's projected growth. Which program should you choose?
A. Enterprise Subscription Agreement
B. Open Value Non-Company-wide
C. Select Plus
D. Enterprise Subscription
Answer: A

NEW QUESTION: 2
Refer to the Exhibit.

How many tracks will be returned to the pool if snapshot 'backup' generation 2 is terminated?
A. 0
B. 1
C. 2
D. 3
Answer: B

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 = ...
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
. . .
B. 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
C. 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
. . .
D. 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
. . .
Answer: A,D

NEW QUESTION: 4
(Multiple choices). The following description about the selection of the distribution column is correct?
A. Join conditions in queries are more suitable as distribution columns.
B. Normally, the primary key of the table is selected as the distribution column.
C. The filter condition of the query is name = 'wang five', so the name column should be used as the distribution column.
D. Columns with fewer values are suitable as distribution columns, which can ensure that the data distribution is relatively concentrated.
Answer: A,B