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NEW QUESTION: 1
You have a Windows Preinstallation Environment (Windows PE) image that uses all of the default configurations. You plan to create a logon script that assigns mapped network drives to user sessions.
You need to identify the drive letter that cannot be used for the mapped network drives.
Which drive letter should you identify?
A. X
B. Z
C. Y
D. W
Answer: D
NEW QUESTION: 2
Which of the following would Pete, a security administrator, do to limit a wireless signal from penetrating the exterior walls?
A. Disable the SSID broadcast
B. Consider antenna placement
C. Implement TKIP encryption
D. Disable WPA
Answer: B
Explanation:
Cinderblock walls, metal cabinets, and other barriers can reduce signal strength significantly. Therefore, antenna placement is critical.
Incorrect Answers:
A. This option deals with encryption, not signal strength.
C. This option would "cloak" the network, not limit its signal strength.
D. This option deals with authentication, not signal strength.
References:
Dulaney, Emmett and Chuck Eastton, CompTIA Security+ Study Guide, 6th Edition, Sybex, Indianapolis, 2014, pp. 172,
173, 177, 183
NEW QUESTION: 3
The realized mean monthly return on the S&P 500 in the 1990's appears to have been substantially different than the mean return in 2000's. The data indicate that assuming equal population variances is not unreasonable.
u = 1990's population mean return and u = 2000's population mean return. The decision made in this
1 2
hypothesis is (assume at the 10% level):
A. not to reject the null hypothesis.
B. that the t value is significant at 0.1 level.
C. to reject the null hypothesis.
Answer: A
Explanation:
Pooled estimate of variance needs to be computed as follows:
2 2 2
S = [(60 - 1)(5.876) + (60 - 1)(4.986) ] / (60 + 60 - 2) = 29.694.
1/2
Now determine the value of t. t = [(0.7 - 1.8) - 0] / [29.964/60 + 29.964/60] = -1.101. We reject null if t >
1 .658 or t -1.658 (t-value: t(118, 0.05) = 1.658). The t value of -1.101 does not fit the rejection criteria. In other words we do not reject the null hypothesis and the t value is not significant at the 0.1 level.