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slavikrds [6]
2 years ago
6

100 POINTS!!!!!!!!

Computers and Technology
2 answers:
dlinn [17]2 years ago
7 0

Answer:

A, B and E

Explanation:

Option A is a correct choice, this is because using an unknown device and connecting it to a network may cause malware to spread on the network or allow a third party unauthorised access to the network.

Option B is also very important because it means that no equipment will be tampered with, keeping it secure is essential.

Option C is not correct as it is simply not possible.

Options D and E are, in my opinion, both correct but I would see how option E is more viable than option D.

Option E is a method hackers use to gain access or data from devices, they do this by making a network that is free and open to everyone and when a device connects, all the data is routed through that network and the person who created it (this is called a man in the middle attack).

Hope this helps!

raketka [301]2 years ago
3 0

Answer:

Option B

Securing Unattended Equipments

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Answer:

piping and instrumentation drawings (P&IDs)

Explanation:

schematic drawings showing all the equipment, pipelines, valves, instruments, and controls needed to operate a piping system

5 0
3 years ago
Please help asap! i’ll give brainliest to the right answer
Lemur [1.5K]

Answer:

import turtle

my_turtle = turtle.Turtle()

my_turtle.speed(0)

my_turtle.pendown()

colours = ["yellow", "blue", "red", "orange", "cyan", "pink", "green", "brown"]

for sides in range(8):

   my_turtle.pencolor(colours[sides])

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Explanation:

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8 0
2 years ago
Lionel wants to find images of colonies of Emperor penguins to use for a school project. Which of the following phrases should h
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7 0
2 years ago
What standard tcp port does the bgp service operate on?
Yanka [14]

Answer:

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3 0
2 years ago
Assume that you are able to do an exhaustive search for the key to an encrypted message at the rate of 100 Million trials per se
pychu [463]

Answer:

8.22 × 10²⁰ years

Explanation:

Given that:

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= 365 × 24 × 60 × 60

= 3.1536 × 10⁷ seconds

Thus, on average, the number of possible keys that is required to check for the decryption should be at least  2¹¹¹ keys.

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