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NEW QUESTION: 1
Martin works as a Software Developer for BlueWell Inc. He creates an ASP.NET application, named App1. During beta testing of App1, he ensures that the actual ASP.NET error messages are displayed whenever errors are encountered. He also ensures that both developers and beta testers see the actual text of the error messages.
During beta testing of App1, Martin performs beta testing of other applications also on the same test server. All the other applications display the ASP.NET error messages. After completing beta testing, Martin promotes the beta test server to a production server. He wants all the applications to display a single, user-friendly error message. Martin also wants to configure App1 and the production server to meet these goals. He removes the customErrors element from the Web.config file for App1. What else will he do to accomplish the required task with minimum administrative effort?
A. In the Machine.config file, add the following element: <customErrors mode="RemoteOnly" />
B. In the Web.config file for App1, add the following element: <customErrors mode="On" />
C. In the App1.config file, add the following element: <customErrors mode="RemoteOnly" />
D. In the App1.config file, add the following element: <customErrors mode="On" />
E. In the Machine.config file, add the following element: <customErrors mode="On" />
F. In the Web.config file for App1, add the following element: <customErrors mode="RemoteOnly" />
Answer: E

NEW QUESTION: 2
Which solution for the IBM DS8880 and IBM Z environments manages complex multisite HA and DR?
A. IBM Copy Services Manager
B. IBM GDPS
C. IBM Storage Insights Pro
D. IBM z/OS Workload Manager
Answer: B

NEW QUESTION: 3
To estimate the average length of their employee's telephone calls, FoneJack, Inc. randomly sampled 25 employee phone calls. If the sample mean was 1.3 minutes and the population can be assumed normal with a standard deviation of 0.3 minutes then a 90% confidence interval for the phone calls is ______.
A. 1.18 m 1.42
B. 1.24 m 1.36
C. 1.2 m 1.4
Answer: C
Explanation:
For a 90% confidence interval we find z(0.05), the cut-off for the top 5% of the normal distribution. Looking up 0.95 in the middle of the table the reading to the row/column values, we get 1.645.
Working with the formula for E (see top right) we get E = 0.1. So, the 90% confidence interval is 1.3 - 0.1 m 1.3 + 0.1 or 1.2 m 1.4.