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NEW QUESTION: 1
The developer wants to write a portable criteria query that will order the orders made by customer James Brown according to increasing quantity. Which one of the below queries correctly accomplishes that task?
A. CriteriaBuilder cb= . . .
CriteriaQuery<order> cq = cb.createquery<order.class>
Root <customer, order> 0 = cq.from(customer.class);
Join <customer, order> 0 = c.Join(customer-.orders);
cq.where (cb.equal(c.get(customer_.name, James Brown)));
cq.orderBy (0.get (order_.quantity));
B. CriteriaBuilder cb= . . .
CriteriaQuery<order> cq = cb.createquery<order.class>
Root <customer, order> 0 = cq.from(customer.class);
Join <customer, order> 0 = c.Join(customer-.orders);
cq.where (cb.equal(c.get(customer_.name, James Brown)));
cq.orderBy (0.get (order_.quantity));
cq.orderBy ("quantity");
C. CriteriaBuilder cb= . . .
CriteriaQuery<order> cq = cb.createquery<order.class>
Root <customer, order> 0 = cq.from(customer.class);
Join <customer, order> 0 = c.Join(customer-.orders);
cq.where (cb.equal(c.get(customer_.name, James Brown))); cq.select(0);
cq.orderBy (0.get (order_.quantity));
D. CriteriaBuilder cb= . . .
CriteriaQuery<order> cq = cb.createquery<order.class>
Root <customer, order> 0 = cq.from(customer.class);
Join <customer, order> 0 = c.Join(customer-.orders);
cq.where (cb.equal(c.get(customer_.name, James Brown)));
cq.orderBy (0.get (order_.quantity));
cq.select(0);
Answer: D

NEW QUESTION: 2
Following a WLC upgrade you want to check what version of code the access point is running. Where can you find this information?
A. GUI > Wireless > Access point > General
B. GUI > Monitor > Software Version
C. CLI > show running configuration
D. CLI > show sysinfo
Answer: C

NEW QUESTION: 3
You work as a database administrator at ABC.com. You study the exhibit carefully.
Exhibit: You have to generate a report that displays the promo name and start date for all promos that started after the last promo in the 'INTERNET' category.

Which query would give you the required output?
A. SELECT promo_name, promo_begin_date FROM promotions
WHERE promo_begin_date > ANY (SELECT promo_begin_date
FROM promotions
WHERE promo_category = 'INTERNET';
B. SELECT promo_name, promo_begin_date FROM promotions
WHERE promo_begin_date > ALL (SELECT MAX(promo_begin_date)
FROM promotions) AND
promo_category = 'INTERNET';
C. SELECT promo_name, promo_begin_date FROM promotions
WHERE promo_begin_date > ALL (SELECT promo_begin_date
FROM promotions
WHERE promo_category = 'INTERNET';
D. SELECT promo_name, promo_begin_date FROM promotions
WHERE promo_begin_date IN (SELECT promo_begin_date
FROM promotions)
WHERE promo_category = 'INTERNET';
Answer: C