Complete Question:
A step commonly used for Internet vulnerability assessment includes __________, which occurs when the penetration test engine is unleashed at the scheduled time using the planned target list and test selection.
Group of answer choices
A. scanning
B. subrogation
C. delegation
D. targeting
Answer:
A. Scanning.
Explanation:
A step commonly used for Internet vulnerability assessment includes scanning, which occurs when the penetration test engine is unleashed at the scheduled time using the planned target list and test selection.
Internet vulnerability assessment can be defined as the process of defining, analyzing, measuring, identifying, classification and prioritization of vulnerabilities in network resources used to access the internet. The main purpose of an internet vulnerability assessment is to provide necessary informations or data about the threats being posed to an individual or organization through the use of a scanning program or system such as a network security scanner
Answer:
The Boolean value returned by that expression will be True
Explanation:
We have two logical statements in that expression:
Expression 1: (10 >= 5*2)
This can be read as: is 10 greater than or equal to 5 multipled by 2. This evaluates to true as 10 is equal to 5 * 2. Hence expression 1 returns true
Expression 2: (10 <= 5*2)
This can be read as: is 10 less than or equal to 5 multiplied by 2. This also evaluates to true as 10 is equal to 5*2. Hence expression 2 returns true.
Now between this two expression is the and operator which evaluates to true if and only if both logical expressions returnes true.
True and True ==> True
Since Expression 1 ==> True and Expression 2 ==> True
This means Expression 1 and Expression 2 ==> True which is the Boolean value returned by the statement
Answer:
2.342m
Explanation:
Given
Time = 0.5 s
Height of Window = 2m
Because the pot was in view for a total of 0.5 seconds, we can assume that it took the cat 0.25 seconds to go from the bottom of the window to the top
Using this equation of motion
S = ut - ½gt²
Where s = 2
u = initial velocity = ?
t = 0.25
g = 9.8
So, we have.
2 = u * 0.25 - ½ * 9.8 * 0.25²
2 = 0.25u - 0.30625
2 + 0.30625 = 0.25u
2.30625 = 0.25u
u = 2.30625/0.25
u = 9.225 m/s ------------ the speed at the bottom of the pot
Using
v² = u² + 2gs to calculate the height above the window
Where v = final velocity = 0
u = 9.225
g = 9.8
S = height above the window
So, we have
0² = 9.225² - 2 * 9.8 * s
0 = 85.100625 - 19.6s
-85.100625 = -19.6s
S = -85.100625/19.6
S = 4.342
If 4.342m is the height above the window and the window is 2m high
Then 4.342 - 2 is the distance above the window
4.342 - 2 = 2.342m
Answer:
algorithms for finding the area
Explanation:
you need algorithms to find out any computer input information.
Answer: See Explanation
Explanation:
The uses of ICT in health department include:
1. ICT helps in the improvement of the safety and the satisfaction of patients as new technologies are being developed to endure that patients are treated faster and their chance of survival increase.
2. ICT helps in looking for prevention measures which will be used to eradicate diseases.
3. ICT helps in the storage of medical data electronically. This will help in the easy retrieval of information.
4. ICT helps in the spread of information and also ensures distant consultation which are essential to achieving health related goals. e.g telemedicine.
5. ICT helps in the easy and fast spread of information and also facilitates cooperation and enhances teamwork among the health workers.
6. ICT brings about efficiency and effectiveness of administrative systems.
