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NEW QUESTION: 1
An administrator is investigating a system that may potentially be compromised and sees the following log entries on the router.
*Jul 15 14:47:29.779: %Router1: list 101 permitted TCP 192.10.3.204(57222) (FastEthernet 0/3) ->
10.10.1.5 (6667), 3 packets.
*Jul 15 14:47:38.779: %Router1: list 101 permitted TCP 192.10.3.204(57222) (FastEthernet 0/3) ->
10.10.1.5 (6667), 6 packets.
*Jul 15 14:47:45.779: %Router1: list 101 permitted TCP 192.10.3.204(57222) (FastEthernet 0/3) ->
10.10.1.5 (6667), 8 packets.
Which of the following BEST describes the compromised system?
A. It is being used in a man-in-the-middle attack
B. It is participating in a botnet
C. It is an ARP poisoning attack
D. It is running a rogue web server
Answer: B

NEW QUESTION: 2
DRAG DROP


Answer:
Explanation:

NEW QUESTION: 3
A security analyst needs to implement an MDM solution for BYOD users that will allow the company to retain control over company emails residing on the devices and limit data exfiltration that might occur if the devices are lost or stolen. Which of the following would BEST meet these requirements? (Select TWO).
A. Containerization
B. Application whitelisting
C. Network usage rules
D. Full-device encryption
E. Remote control
F. Geofencing
Answer: A,D

NEW QUESTION: 4
Suppose you have two assets, A and B Over the past 3 periods, A has returned 8%, 2%, and 6%, while B has returned 11%, -5%, and 20%. What is the return covariance between assets A and B?
A. 10.21%%
B. 19.79%%
C. 0.0%%
D. 31.38%%
Answer: B
Explanation:
Explanation/Reference:
Explanation:
First we must find the expected returns for A and B These are 5.33% and 8.67%. Second, we find the difference between each observation and the average: (8% - 5.33%), (2% - 5.33%), and (6% - 5.33%) for A, and (11% - 8.67%), (-5% - 8.67%), and (20% - 8.67%) for B Next, we multiply these together and sum them: (8% - 5.33%)*(11% - 8.67%) + (2% - 5.33%)*(-5% - 8.67%) + (6% - 5.33%)*(20% - 8.67%). The sum of these is 59.33%. The covariance is the probability weighted average of these cross products, so we divide by 3 to get 19.78%%. Note, we could have divided each cross product by 3 rather than the sum of the cross products. If the observations did not have the same probability or frequency, we would need to treat each cross product separately rather than divide at the end.