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NEW QUESTION: 1
Ein Netzwerkadministrator hat die Ordner mehrerer Benutzer auf einen anderen Server verschoben. Nach der Zuordnung zum neuen freigegebenen Laufwerk können die Benutzer auf die Dateien der anderen zugreifen. Welche der folgenden Ursachen könnte dieses Problem verursachen?
A. Administratorrechte
B. Netzwerksegmentierung
C. Dateifragmentierung
D. Dateiberechtigungen
Answer: D

NEW QUESTION: 2

For each of the following statements, select Yes if the statement is true. Otherwise, select No. NOTE: Each correct selection is worth one point.

Answer:
Explanation:

NEW QUESTION: 3
次の監査手法のうち、監査の分析プロセスを具現化しながら制御設計を評価するために使用されるものはどれですか。
A. プロセスの説明。
B. ウォークスルー。
C. リスクと管理のマトリックス。
D. フローチャート。
Answer: C

NEW QUESTION: 4
A researcher has a sample of 700 observations from a population whose standard deviation is known to be
1,235.6. The mean of the sample is calculated to be 219.2. The null hypothesis is stated as Ho: mean
150 and the alternative is non-directional. The p-value in this case equals ________.
A. 10.16%
B. 13.88%
C. 6.94%
D. 12.82%
Answer: B
Explanation:
Explanation/Reference:
Explanation:
The z-statistic under the null equals (219.2 - 150)/(1235.6/(700

HOT EXAMS

.5)) = 1.48. Since the alternative is non- directional, you must use a two-tailed test and ask the question, "What is the probability of observing a z- statistic at least as large in magnitude as 1.48?" The answer to this will give you the p- value for this sample under the two-tailed test. The right-tailed probability of observing a z-statistic which is at least as big as 1.48 equals 1.0 - 0.9306 = 0.0694 = 6.94%. The left-tailed probability of observing a z-statistic which is at least as small as - 1.48 also equals 6.94% since the standard normal distribution is symmetrical about zero. Therefore, the p-value of this two-tailed test in this sample equals 6.94%*2 = 13.88%.

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