Answer: A it is true.
Explanation: It is true under Subpart C—Information Security Responsibilities for Employees who Manage or Use Federal Information Systems
Number 3 and 4
(3) Program and functional managers must receive training in information security basics; management and implementation level training in security planning and system/application security management; and management and implementation level training in system/application life cycle management, risk management, and contingency planning.
(4) Chief Information Officers (CIOs), IT security program managers, auditors, and other security-oriented personnel (e.g., system and network administrators, and system/application security officers) must receive training in information security basics and broad training in security planning, system and application security management, system/application life cycle management, risk management, and contingency planning.
Answer: In a clean plastic or metal container with a tightly sealed lid
Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
<em>4.3 Print count</em>
<em>4.4 count = 0</em>
End the outer loop
4.5 End i loop
End the algorithm
5. End
Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
