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NEW QUESTION: 1
A 45-year-old female with sarcoidosis has pulmonary function tests annually. The following data have been recorded:

A pulmonary function technologist should conclude that
A. The patient has a normal rate of loss for FEV1.
B. The patient has developed mild obstructive airway disease.
C. The patient's restrictive component has worsened.
D. Incorrect predicted values have been used.
Answer: D

NEW QUESTION: 2
Which of the following is true for Single loss expectancy (SLE), Annual rate of occurrence (ARO), and Annual loss expectancy (ALE)?
A. Explanation:
A quantitative risk assessment quantifies risk in terms of numbers such as dollar values. This involves gathering data and then entering it into standard formulas. The results can help in identifying the priority of risks. These results are also used to determine the effectiveness of controls. Some of the terms associated with quantitative risk assessments are : Single loss expectancy (SLE)-It refers to the total loss expected from a single incident. This incident can occur when vulnerability is being exploited by threat. The loss is expressed as a dollar value such as $1,000. It includes the value of data, software, and hardware. SLE = Asset value * Exposure factor Annual rate of occurrence (ARO)-It refers to the number of times expected for an incident to occur in a year. If an incident occurred twice a month in the past year, the ARO is 24. Assuming nothing changes, it is likely that it will occur 24 times next year. Annual loss expectancy (ALE)-It is the expected loss for a year. ALE is calculated by multiplying SLE with ARO. Because SLE is a given in a dollar value, ALE is also given in a dollar value. For example, if the SLE is $1,000 and the ARO is 24, the ALE is $24,000. ALE = SLE * ARO Safeguard value-This is the cost of a control. Controls are used to mitigate risk. For example, antivirus software of an average cost of $50 for each computer. If there are 50 computers, the safeguard value is $2,500.
B. ALE= ARO*SLE
C. ALE= ARO/SLE
D. ARO= SLE/ALE
E. ARO= ALE*SLE
Answer: A,B
Explanation:
A, and B are incorrect. These are wrong formulas and are not used in quantitative risk assessment.

NEW QUESTION: 3
How does TCP differ from UDP? (Choose two.)
A. TCP provides best effort delivery.
B. TCP provides synchronized communication.
C. TCP segments are essentially datagrams.
D. TCP provides sequence numbering of packets.
E. TCP uses broadcast delivery.
Answer: B,D
Explanation:
Because TCP is a connection-oriented protocol responsible for ensuring the transfer of a datagram from the source to
destination machine (end-to-end communications), TCP must receive communications messages from the destination
machine to acknowledge receipt of the datagram. The term virtual circuit is usually used to refer to the handshaking
that goes on between the two end machines, most of which are simple acknowledgment messages (either
confirmation of receipt or a failure code) and datagram sequence numbers.
Rather than impose a state within the network to support the connection, TCP uses synchronized state between the
two endpoints. This synchronized state is set up as part of an initial connection process, so TCP can be regarded as a
connection-oriented protocol. Much of the protocol design is intended to ensure that each local state transition is
communicated to, and acknowledged by, the remote party.
Reference: http://en.wikibooks.org/wiki/Communication_Networks/TCP_and_UDP_Protocols

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