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NEW QUESTION: 1
An administrator has deployed an application that uses 30 unique SQL statements with a maximum data source connection pool size of 10.
What is the recommended prepared statement cache size for the application?
A. 0
B. 1
C. 2
D. 3
Answer: D
Explanation:
To determine the highest value for your cache size to avoid any cache discards, add the number of uniquely prepared statements and callable statements, as determined by the
SQL string, concurrency, and the scroll type, for each application that uses this data source on a particular server. This value is the maximum number of possible statements that can be cached on a given connection over the life of the server. Setting the cache size to this value means that you never have cache discards. In general, configure a larger cache for applications with a greater number of statements.
References:https://www.ibm.com/support/knowledgecenter/en/SSAW57_8.5.5/com.ibm.we bsphere.nd.doc/ae/udat_jdbcdatasorprops.html

NEW QUESTION: 2
An administrator discovers the following log entry on a server:
Nov 12 2013 00:23:45
httpd[2342]: GET/app2/prod/proc/process.php?input=change;cd%20../../../etc;cat%20shadow
Which of the following attacks is being attempted?
A. Cross-site scripting
B. Password attack
C. Buffer overflow
D. Command injection
Answer: B

NEW QUESTION: 3
Heniser Pet Foods는 X.와 Y의 두 가지 제품을 생산합니다.
제품 X와 Y는 각각 미화 30 달러와 50 달러입니다. 각 제품은 재료 A와
B. 제품은 6 파운드의 재료 A와 12 파운드의 재료 B를 사용합니다. 제품 Y는 12 파운드의 재료 A와 8 파운드의 재료 B를 사용합니다. 회사는 1,200 파운드의 재료 A와 1,760 파운드의 재료 B 만 구입할 수 있습니다. 제조 할 제품의 최적 혼합은 다음과 같습니다.
A. 120 단위 및 40 단위 Y.
B. 146 단위 및 Y 단위 0 단위.
C. 0 단위 X 및 100 단위 Y.
D. 40 단위의 X 및 120 단위의 Y.
Answer: A
Explanation:
Linear programming is a technique used to maximize a contribution margin function or to minimize a cost function, subject to constraints such as scarce resources or minimum/maximum levels of production. Thus, linear programming is often used for planning resource allocations. In this problem, the equation to be maximized, called the objective function, is: U3 $30)K+ $50Y. This equation is to be maximized subject to the constraints on materials. The two constraint functions are:
Material A: 6X + 12Y < 1,200 Material B: I 2X + 8Y < 1,760
One way to solve this problem is to graph the constraint lines and determine the feasible area. The optimal production level is at an extreme point within the feasible area. The graph shows that a production level of 120 units of and 40 units of Y is a feasible production level that maximizes the contribution margin.