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NEW QUESTION: 1
HOTSPOT
Match each of the following entities that Veeam Backup & Replication uses to manage tapes with its definition.
Answer:
Explanation:
NEW QUESTION: 2
DRAG DROP
You develop a Windows Communication Foundation (WCF) service. The service implements the IRegistrationService interface in a class named RegistrationService.
You need to configure the service to use file-less activation.
How should you complete the relevant code? (To answer, drag the appropriate code segments to the correct location or locations in the answer area. Each code segment may be used once, more than once, or not at all. You may need to drag the split bar between panes or scroll to view content.)
Answer:
Explanation:
NEW QUESTION: 3
A. Option D
B. Option A
C. Option B
D. Option C
Answer: D
Explanation:
The scalability with high availability configuration ensures that there is a messaging engine for each server in a cluster, and that each messaging engine has a failover location.
Reference:http://www01.ibm.com/support/knowledgecenter/SSCKBL_8.5.5/com.ibm.websphere.nd.doc/ae/cjt1004_.html(second paragraph)
NEW QUESTION: 4
The sponsors of a well-known charity came up with a unique idea to attract wealthy patrons to the
$ 500 a plate dinner. After the dinner, it was announced that each patron attending could buy a set of 20 tickets for the gaming tables. The chance of winning a prize for each of the 20 plays is 50-50. If you bought a set of 20 tickets, what is the chance that you will win 15 or more prizes?
A. 0.250
B. 0.021
C. 0.006
Answer: B
Explanation:
This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by:
r (n-r)
n!(p )[q ]/r!(n-r)!. Here n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore we have
15 5
P(15) = 20!(0.5 )(0.5 )/15!5! = 0.0148
16 4
P(16) = 20!(0.5 )(0.5 )/16!4! = 0.0046 P(17) = 0.0011 P(18) = 0.0002 P(19) = 0.00002 P(20) = 0.000001
The sum adds up to 0.0207.
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