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NEW QUESTION: 1
Bei der Ermittlung einer früheren Projektabschlusszeit, die durch Zahlung einer Prämie für den vorzeitigen Abschluss erzielt werden soll, sollten folgende Aktivitäten ausgewählt werden:
A. Die Summe der Aktivitätszeiten ist die kürzeste.
B. Dies ergibt die längstmögliche Ausführungszeit.
C. Die Summe der Pausenzeiten ist die kürzeste.
D. Diese haben keine Nachlaufzeit.
Answer: D
Explanation:
Erläuterung:
Die Aktivitätszeit eines kritischen Pfads ist länger als bei jedem anderen Pfad durch das Netzwerk. Dieser Pfad ist wichtig, da, wenn alles wie geplant verläuft, seine Länge die kürzestmögliche Ausführungszeit für das Gesamtprojekt ergibt. Aktivitäten auf dem kritischen Pfad werden zu Kandidaten für Abstürze, d. H. Zur Verkürzung ihrer Zeit durch Zahlung einer Prämie für die vorzeitige Fertigstellung. Aktivitäten auf dem kritischen Pfad haben eine Nachlaufzeit von Null, und umgekehrt befinden sich Aktivitäten mit einer Nachlaufzeit von Null auf einem kritischen Pfad. Durch sukzessives Entspannen von Aktivitäten auf einem kritischen Pfad kann eine Kurve erhalten werden, die die Gesamtkosten des Projekts im Verhältnis zur Zeit anzeigt.

NEW QUESTION: 2
Which CSS3 code fragment uses a pseudo-element?
A. div+p {font-weight: bold;}
B. div>p {font-weight: bold;}
C. p. first-letter {font-weight: bold;}
D. p: : first-letter {font-weight: bold;}
Answer: D
Explanation:
Explanation/Reference:
References:
http://www.html5code.nl/css3-tutorials/css3-tutorial-css3-selectors/

NEW QUESTION: 3
Examine the data in the CUST_NAME column of the CUSTOMERS table.
CUST_NAME
---------------------
Lex De Haan
Renske Ladwig
Jose Manuel Urman
Jason Mallin
You want to extract only those customer names that have three names and display the * symbol in place of the first name as follows:
CUST NAME
---------------------
*** De Haan
**** Manuel Urman
Which two queries give the required output? (Choose two.)
A. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')), LENGTH(cust_name)- INSTR(cust_name, ''), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ', -
1, -2)<>0;
B. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')), LENGTH(cust_name), '*')
"CUST NAME" FROM customersWHERE INSTR(cust_name, ' ', -1, 2)<>0;
C. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')), LENGTH(cust_name), '*')
"CUST NAME" FROM customersWHERE INSTR(cust_name, ' ', 1, 2)<>0;
D. SELECT LPAD(SUBSTR(cust_name, INSTR(cust_name, ' ')), LENGTH(cust_name)- INSTR(cust_name, ' '), '*') "CUST NAME"FROM customersWHERE INSTR(cust_name, ' ',
1, 2)<>0 ;
Answer: B,C