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NEW QUESTION: 1
Es wird ein Nachweis geliefert, der die genehmigten Geschäftsanforderungen erfüllt. Der Kunde identifiziert jedoch einige nicht zusammenhängende Probleme, die zu einer Verzögerung der endgültigen Produktgenehmigung führen und sich auf den Projektabschluss auswirken. Was sollte der Projektmanager tun?
A. Lösen Sie die nicht zusammenhängenden Probleme, erhalten Sie Akzeptanz und schließen Sie das Projekt offiziell ab
B. Folgen Sie dem Eskalationsprozess, entwickeln Sie den Übergabebericht und schließen Sie das Projekt offiziell ab
C. Besprechen Sie die Auswirkungen mit dem Kunden, holen Sie die Akzeptanz ein und schließen Sie das Projekt offiziell ab
D. Dokumentieren Sie die nicht zusammenhängenden Probleme. Entwickeln Sie den Übergabebericht und schließen Sie das Projekt offiziell ab
Answer: D
NEW QUESTION: 2
Which of the following statements about RAID 6 is incorrect?
A. RAID 6 requires two-level parity.
B. RAID 6 requires at least three member disks.
C. Frequently used RAID 6 technologies include RAID 6 P+Q and RAID 6 DP.
D. RAID 6 can recover data after two member disks fail.
Answer: B
NEW QUESTION: 3
An engineer is working on a Cisco VCS Control routing configuration and wants users to be able to dial ccnpcollab and have calls routed to [email protected]. Which option achieves this aim?
A. access rules
B. transforms
C. call policy
D. search rules
Answer: C
NEW QUESTION: 4
Examine the data in the CUST_NAMEcolumn of the CUSTOMERStable.
CUST_NAME
-------------------
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikkilineni
Julia Nayer
You need to display customers' second names where the second name starts with "Mc" or "MC".
Which query gives the required output?
SELECT SUBSTR(cust_name, INSTR (cust_name, ' ')+1)
A. FROM customers
WHERE INITCAP(SUBSTR(cust_name, INSTR (cust_name, ' ')+1))
LIKE 'Mc%';
SELECT SUBSTR(cust_name, INSTR (cust_name, ' ')+1)
B. FROM customers
WHERE INITCAP(SUBSTR(cust_name, INSTR (cust_name, ' ')+1))
'Mc';
SELECT SUBSTR(cust_name, INSTR (cust_name, ' ')+1)
C. FROM customers
WHERE INITCAP(SUBSTR(cust_name, INSTR (cust_name, ' ')+1))
INITCAP ('MC%');
D. FROM customers
WHERE SUBSTR(cust_name, INSTR (cust_name, ' ')+1)
LIKE INITCAP ('MC%');
SELECT SUBSTR(cust_name, INSTR (cust_name, ' ')+1)
Answer: A