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NEW QUESTION: 1
A company requires that laborers record the actual time they spent on a work order. The labor hours recorded are not automatically approved and the company does not want the laborer to have access to the Work Order Tracking application.
Which application would the laborers use?
A. Invoicing
B. Time Sheets
C. Labor Reporting
D. Activities and Tasks
Answer: C
Explanation:
Explanation/Reference:
Reference: http://www.ibm.com/support/knowledgecenter/SSLKT6_7.5.0/com.ibm.mam.doc/gp_finmgr/ c_ctr_rprt_lab_use.html
NEW QUESTION: 2
注：この質問は、同じシナリオを提示する一連の質問の一部です。 シリーズの各質問には、記載された目標を達成する可能性のある独自のソリューションが含まれています。 一部の質問セットには複数の正しい解決策があり、他の質問セットには正しい解決策がない場合があります。
このシナリオで質問に回答すると、その質問に戻ることはできません。 その結果、これらの質問はレビュー画面に表示されません。
Azure Cosmos DBデータベースにSalesという名前のコンテナーがあります。 Salesには120 GBのデータがあります。 Salesの各エントリには、次の構造があります。
パーティションキーはOrderId属性に設定されます。
ユーザーは、ProductNameでデータを取得するクエリを実行すると、クエリの完了に予想よりも長い時間がかかると報告しています。
問題のあるクエリの実行にかかる時間を短縮する必要があります。
解決策：データベースの要求単位（RU）を増やします。
これは目標を達成していますか？
A. いいえ
B. はい
Answer: B
Explanation:
Explanation
To scale the provisioned throughput for your application, you can increase or decrease the number of RUs at any time.
Note: The cost of all database operations is normalized by Azure Cosmos DB and is expressed by Request Units (or RUs, for short). You can think of RUs per second as the currency for throughput. RUs per second is a rate-based currency. It abstracts the system resources such as CPU, IOPS, and memory that are required to perform the database operations supported by Azure Cosmos DB.
Reference:
https://docs.microsoft.com/en-us/azure/cosmos-db/request-units
NEW QUESTION: 3
Given the following D-spec:
DName+++++++++++ETDsFrom+++To/L+++IDc.Keywords++++++++++++++++++++++
d EndOfMonth s d
Which of the following expressions will set EndOfMonth to the last day of the previous month?
A. EndOfMonth = %subdt(%date() : %day(%date()));
B. EndOfMonth = %date - %days(%subdt(%date() : *d)) - %days(1);
C. EndOfMonth = %date() - %days(%subdt(%date() : *d));
D. EndOfMonth = %subdt(%date() : %day(%date())) - %days(1);
Answer: C