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  • Exam Code: NSE6_FSR-7.3
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NEW QUESTION: 1
Drag and drop the descriptions from the left onto correct application deployment models on the right.

Answer:
Explanation:

Explanation
1 - B, 2 - E, 3 - C, 4 - D, 5 - A


NEW QUESTION: 2
Note: This question is part of a series of questions that use the same or similar answer choices. An answer choice may be correct for more than one question in the series. Each question is independent of the other questions in this series. Information and details provided in a question apply only to that question.
You have a database that contains tables named Customer_CRMSystem and Customer_HRSystem. Both tables use the following structure:

The tables include the following records:
Customer_CRMSystem

Customer_HRSystem

Records that contain null values for CustomerCode can be uniquely identified by CustomerName.
You need to display a Cartesian product, combining both tables.
Which Transact-SQL statement should you run?


A. Option C
B. Option G
C. Option H
D. Option B
E. Option E
F. Option D
G. Option F
H. Option A
Answer: B
Explanation:
Explanation
A cross join that does not have a WHERE clause produces the Cartesian product of the tables involved in the join. The size of a Cartesian product result set is the number of rows in the first table multiplied by the number of rows in the second table.
References: https://technet.microsoft.com/en-us/library/ms190690(v=sql.105).aspx

NEW QUESTION: 3
展示を表示し、ORDERSテーブルとCUSTOMERSテーブルの構造を調べます。

CUST_LAST_NAME列に値Robertsを持つ顧客は1人だけです。
CUST_LAST_NAMEがRobertsでCREDIT_LIMITが600である顧客のORDERSテーブルに行を追加するには、どのINSERTステートメントを使用する必要がありますか?
A. INSERT INTO ordersVALUES(1、'10 -mar-2007 '、' direct '、(SELECT customer_idFROM CustomersWHERE cust_last_name =' Roberts 'ANDcredit_limit = 600)、1000);
B. INSERT INTO注文(order_id、order_date、order_mode、(SELECT customer_idFROM CustomersWHERE cust_last_name = 'Roberts' ANDcredit_limit = 600)、order_total)VALUES(1、'10 -mar-2007 '、' direct '、&& customer_id、1000) ;
C. INSERT INTO(SELECT o.order_id、o.order_date、o.order_mode、c.customer_id、o.order_totalFROM orders o、Customers cWHERE o.customer_id = c.customer_idAND c.cust_last_name = 'Roberts' ANDc.credit_limit = 600)VALUES(1、'10 -mar-2007 '、' direct '、(SELECT customer_idFROM CustomersWHERE cust_last_name =' Roberts 'ANDcredit_limit = 600)、1000);
D. INSERT INTO注文(order_id、order_date、order_mode、(SELECT customer_idFROM CustomersWHERE cust_last_name = 'Roberts' ANDcredit_limit = 600)、order_total)VALUES(1、'10 -mar-2007 '、' direct '、&customer_id、1000) ;
Answer: A

NEW QUESTION: 4
Which two parameters are needed to do route peering with a Layer 4-Layer 7 service device in Cisco ACI?
(Choose two.)
A. Under the device interface, the path must be selected.
B. Under the device interface, the L3 Out must be selected.
C. The connector type must be set to "Layer 2 peering".
D. A router ID must be assigned under the Layer 4-layer 7 device
E. A valid router configuration must be set under the service graph template.
Answer: D,E
Explanation:
Explanation/Reference:
Reference: https://www.cisco.com/c/en/us/td/docs/switches/datacenter/aci/apic/sw/1-x/L4- L7_Services_Deployment/guide/b_L4L7_Deploy_ver201/b_L4L7_Deploy_ver201_chapter_01000.html

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