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HACTEHA [7]
2 years ago
15

3.3 Code Practice: Question 1

Computers and Technology
1 answer:
NeX [460]2 years ago
6 0

Answer:

Question1:

day = int(input("Enter in Numeric Today's day:"))

if (x == 15 or x == 30):

  print("You have a payday today!")

if (x != 15 and x != 30):

  print("Its is not certainly a payday today but dont feel bad.")

Question 2:

red = int(input("Enter the value for red: "))

green = int(input("Enter the value for green: "))

blue = int(input("Enter the value for blue: "))

if (red > 255 or red < 0):

  print("The value of Red is not correct.")

if (green > 255 or green < 0):

  print("The value of Green is not correct.")

if (blue > 255 or blue < 0):

  print("The value of Blue is not correct.")

Explanation:

Please check the answer section.

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How do i write this in c++ = scnr.nextInt();<br> please help
LenaWriter [7]
Int whatever = scnr.nextInt();



Although usually when a class has a next* member function, it usually needs to check that there IS a next, that you haven't reached the end. Without knowing the class that scnr was instantiated from, I can't guess.
7 0
3 years ago
__________________________would be involved in projects such as implementing security measures and software to protect systems a
defon

<u>Network Administrator</u><u>/Network Engineer</u> would be involved in projects such as implementing security measures and software to protect systems and information infrastructure, including firewalls and data-encryption programs.

<h3>What does an administrator of a network do?</h3>

The  network administrator is known to be a person who is involved in the day-to-day management of these networks is the responsibility of network and computer system administrators.

They function by helping to plan, set up, and provide maintenance for a company's computer systems, etc., for data communication.

Therefore, <u>Network Administrator</u><u>/Network Engineer</u> would be involved in projects such as implementing security measures and software to protect systems and information infrastructure, including firewalls and data-encryption programs.

Hence, option D is correct.

Learn more about Network Administrator from

brainly.com/question/5860806
#SPJ1

3 0
10 months ago
Can you fart and burp at the same time?
Rina8888 [55]

Answer:

Yes you can

Explanation:

Although farting and burping at the same time is a very rare phenomenon, it’s very possible. When this happens at the same time it is usually called a Furp.

This occurrence usually happens when there’s a lot of intake of foods which have a large percentage of gas . These gases often need to be expelled through processes such as burping and farting.

3 0
3 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
3 years ago
HELP PLEASE GIVING BRAINLIEST
Phoenix [80]

Answer:

CompTIA A+ there ya go!!!

5 0
3 years ago
Read 2 more answers
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