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NEW QUESTION: 1
どの三つシスコUnified Wireless Network能力が、ラジオResource Management隣人メッセージによって提供される情報を使いますか？（3を選んでください。）
A. 積極的なロード・バランシング
B. 電波放送されて配給すること
C. コントローラ間のモビリティ (つまり、モビリティグループ)
D. 複合型遠い端アクセス・ポイント
E. ダイナミックなチャンネル割当て
F. 危険なAP分類
Answer: B,E,F
NEW QUESTION: 2
技術者がPHIを含むコンピューターハードウェアを処分しています。ドライブは再利用可能でなければなりません。次の方法のうちどれを使用する必要がありますか？
A. デガウス
B. 標準フォーマット
C. 低レベル形式
D. ドライブワイプ
Answer: D
NEW QUESTION: 3
View the Exhibit and examine the structure of the CUSTOMERS table.
Which two tasks would require subqueries or joins to be executed in a single statement? (Choose two.)
A. listing of those customers whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'
B. listing of customers who do not have a credit limit and were born before 1980
C. finding the number of customers, in each city, whose marital status is 'married'
D. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
E. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
Answer: A,D
Explanation:
Describe the Types of Problems That the Subqueries Can Solve
There are many situations where you will need the result of one query as the input for another.
Use of a Subquery Result Set for Comparison Purposes
Which employees have a salary that is less than the average salary? This could be answered by
two statements, or by a single statement with a subquery. The following example uses two
statements:
select avg(salary) from employees;
select last_name from employees where salary < result_of_previous_query ;
Alternatively, this example uses one statement with a subquery:
select last_name from employees where salary < (select avg(salary)from employees);
In this example, the subquery is used to substitute a value into the WHERE clause of the parent
query: it is returning a single value, used for comparison with the rows retrieved by the parent
query.
The subquery could return a set of rows. For example, you could use the following to find all
departments that do actually have one or more employees assigned to them:
select department_name from departments where department_id in
(select distinct(department_id) from employees);