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NEW QUESTION: 1
Universal Containers sells their subscription products only in whole months. The Users at Universal Containers are sometimes entering values into Start Date and End Date and leaving Subscription Term blank. They are finding that subscription pricing is not working as they would expect because the dates they are choosing do not equate to exact months. This leads to pricing that may be a few cents or even a few dollars off. The Admin would like to find a solution that does not require forcing the users to use Subscription Term. What should be done so that the pricing is rounded to the expected value?
A. Create a Price Rule which rounds the Prorate Multiplier to the nearest whole value.
B. In the package settings, set Subscription Prorate Precision to Monthly.
C. On the Quote object, set the Subscription Prorate Precision field to Monthly.
D. Create a Price Rule that takes the user-entered date and calculates a rounded Term.
Answer: B
NEW QUESTION: 2
Which one of the following queries selects the customer whose order has the highest total price?
A. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.cteateQuery(Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
B. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.createQuery(Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = sq.from(Order.class);
sq. select (ci: . max ( subo . get (Order_ . Total Price) ) ) ;
cq.where(sq.all(o.gei(Order_.totalPrice)));
C. CriteriaBuilder cb = ...
Criteria Query <Customer> cq = cb.create Query (Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
D. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.createquery(customer.class)
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
Answer: D
NEW QUESTION: 3
R1# show running-config
interface Loopback0
description ***Loopback***
ip address 192.168.1.1 255.255.255.255
ip ospf 1 area 0
!
interface Ethernet0/0
description **Connected to R1-LAN**
ip address 10.10.110.1 255.255.255.0
ip ospf 1 area 0
!
interface Ethernet0/1
description **Connected to L2SW**
ip address 10.10.230.1 255.255.255.0
ip ospf hello-interval 25
ip ospf 1 area 0
!
router ospf 1
log-adjacency-changes
R2# show running-config
R2
!
interface Loopback0
description **Loopback**
ip address 192.168.2.2 255.255.255.255
ip ospf 2 area 0
!
interface Ethernet0/0
description **Connected to R2-LAN**
ip address 10.10.120.1 255.255.255.0
ip ospf 2 area 0
!
interface Ethernet0/1
description **Connected to L2SW**
ip address 10.10.230.2 255.255.255.0
ip ospf 2 area 0
!
router ospf 2
log-adjacency-changes
R3# show running-config
R3
username R6 password CISCO36
!
interface Loopback0
description **Loopback**
ip address 192.168.3.3 255.255.255.255
ip ospf 3 area 0
!
interface Ethernet0/0
description **Connected to L2SW**
ip address 10.10.230.3 255.255.255.0
ip ospf 3 area 0
!
interface Serial1/0
description **Connected to R4-Branch1 office**
ip address 10.10.240.1 255.255.255.252
encapsulation ppp
ip ospf 3 area 0
!
interface Serial1/1
description **Connected to R5-Branch2 office**
ip address 10.10.240.5 255.255.255.252
encapsulation ppp
ip ospf hello-interval 50
ip ospf 3 area 0
!
interface Serial1/2
description **Connected to R6-Branch3 office**
ip address 10.10.240.9 255.255.255.252
encapsulation ppp
ip ospf 3 area 0
ppp authentication chap
!
router ospf 3
router-id 192.168.3.3
!
R4# show running-config
R4
!
interface Loopback0
description **Loopback**
ip address 192.168.4.4 255.255.255.255
ip ospf 4 area 2
!
interface Ethernet0/0
ip address 172.16.113.1 255.255.255.0
ip ospf 4 area 2
!
interface Serial1/0
description **Connected to R3-Main Branch office**
ip address 10.10.240.2 255.255.255.252
encapsulation ppp
ip ospf 4 area 2
!
router ospf 4
log-adjacency-changes
R5# show running-config
R5
!
interface Loopback0
description **Loopback**
ip address 192.168.5.5 255.255.255.255
ip ospf 5 area 0
!
interface Ethernet0/0
ip address 172.16.114.1 255.255.255.0
ip ospf 5 area 0
!
interface Serial1/0
description **Connected to R3-Main Branch office**
ip address 10.10.240.6 255.255.255.252
encapsulation ppp
ip ospf 5 area 0
!
router ospf 5
log-adjacency-changes
R6# show running-config
R6
username R3 password CISCO36
!
interface Loopback0
description **Loopback**
ip address 192.168.6.6 255.255.255.255
ip ospf 6 area 0
!
interface Ethernet0/0
ip address 172.16.115.1 255.255.255.0
ip ospf 6 area 0
!
interface Serial1/0
description **Connected to R3-Main Branch office**
ip address 10.10.240.10 255.255.255.252
encapsulation ppp
ip ospf 6 area 0
ppp authentication chap
!
router ospf 6
router-id 192.168.3.3
!
A. Option B
B. Option C
C. Option A
D. Option D
Answer: A
NEW QUESTION: 4
During a high traffic portion of the day, one of your relational databases crashes, but the replica is never promoted to a master. You want to avoid this in the future. What should you do?
A. Use a different database.
B. Implement routinely scheduled failovers of your databases.
C. Create snapshots of your database more regularly.
D. Choose larger instances for your database.
Answer: D