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miskamm [114]
3 years ago
8

In the 5 Code of Federal Regulations (C.F.R.), it is recommended that an individual has security awareness training before s/he

can access information. The C.F.R is unusual in that it requires all users to receive broad training in system/application life cycle management, security planning and system/application security management, risk management, and contingency planning.A. TrueB. false
Engineering
2 answers:
GarryVolchara [31]3 years ago
6 0

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.

Helen [10]3 years ago
5 0

Answer:

B. False

Explanation:

The Code of Federal Regulations (C.F.R.) also called the administrative law refers to the codes of conducts that governs and sets the boundaries for administrative agencies in the United States.

The C.F.R. code stipulates that every individual who wants to access the Federal Information System be exposed to a certain basic level of security awareness before doing so. However, the C.F.R. does not stipulate that all users undergo broad training in system/application life cycle management, security planning and system/application security management, risk management, and contingency planning.

According to the C.F.R. code, training is made available to individuals based on their roles and responsibilities; Executives receive a basic security training, Program managers receive management training as well as basic security training, Chief Information Officers & other security-oriented personnels are the ones that receive broad training in system/application life cycle management, security planning and system/application security management, risk management, and contingency planning.

The C.F.R. code also stipulates that new employees be introduced & acquainted with security training depending on their roles/positions before granting them access to the systems, that current employees be refreshed often with security training and to make security training available to employees when there is any significant change in the agency's information system procedure or if an employee is given a new role that demands additional training.

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Why data structure is important
Readme [11.4K]

Answer:

Data structures are important as it allows the user to insert, update, arrange, rearrange, delete, and retrieve data in an efficient manner, from the database. And to accomplish the said tasks algorithms are used. It is used to manipulate the stored data within the database in the required manner. …

Explanation:

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8 0
2 years ago
The current through a 10-mH inductor is 10e−t∕2 A. Find the voltage and the power at t = 8 s.
NNADVOKAT [17]

Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L\frac{di(t)}{dt}             -----------(i)

Where;

L = inductance of the inductor

i(t) = current through the inductor at a given time

t = time for the flow of current

From the question:

i(t) = 10e^{-t/2}A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = (10*10^{-3})\frac{d(10e^{-t/2})}{dt}

Solve the differential

v(t) = (10*10^{-3})\frac{-1*10}{2} (e^{-t/2})

v(t) = -0.05 e^{-t/2}

At t = 8s

v(t) = v(8) = -0.05 e^{-8/2}

v(t) = v(8) = -0.05 e^{-4}

v(t) = -0.05 x 0.223

v(t) = -0.01116V

(b) To get the power, we use the following relation:

p(t) = i(t) x v(t)

Power at t = 8

p(8) = i(8) x v(8)

i(8) = i(t = 8) = 10e^{-8/2}

i(8) = 10e^{-4}

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

7 0
3 years ago
Read 2 more answers
Two technicians are discussing relays. Technician A says that relays can fail because the relay winding is open. Technician B sa
hram777 [196]

Technician A says that relays can fail because the relay winding is open. Technician A is correct.

<h3>What is winding?</h3>

A single turn of the wound material, which is material wrapped or coiled around an object.

Poor contact alignment and open coils can also cause relays to malfunction.

The most important aspect impacting relay dependability is choosing the right relay type for a particular application. When incorporating them into circuits, several subpar design techniques are employed.

Hence, technician A is correct.

To learn more about the winding refer;

brainly.com/question/23369600

#SPJ1

4 0
1 year ago
TWO SENTENCES!!! What is something that you have used today that was designed by an engineer? What parts were designed by an eng
svetlana [45]
Cars, houses, Devices. :)
6 0
3 years ago
Read 2 more answers
A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar
jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading  114.06 m^3/d/m

Explanation:

calculation for single clarifier

sewag\  flow Q = \frac{12900}{2} = 6450 m^2/d

surface\  area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2

surface area = 314.16 m^2

volume of tankV  = A\times side\ water\ depth

                             =314.16\times 2 = 628.32m^3

Length\ of\  weir = \pi \times diameter of weir

                       = \pi \times 18 = 56.549 m

overflow rate =v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2

Detention timet_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr

weir loading= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m

6 0
2 years ago
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