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Nat2105 [25]
3 years ago
9

Who plays Roblox and wants to be friends on it

Computers and Technology
shadow Rain
2 years ago
I play Roblox, account: chemical421
2 answers:
qaws [65]3 years ago
4 0
I have roblox and I would like to be friends on it
valentina_108 [34]3 years ago
3 0

Answer:

I have a Roblox account but I don't play.

Explanation:

quite useless answer so sorry

You might be interested in
In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the r
Zielflug [23.3K]

Answer: 10 students

Explanation:

Students younger than 10 = 50%

Students that are 10years old = 1/20 = 1/20 × 100 = 5%

Students that are older than 10 but younger than 12 = 1/10= 1/10 × 100 = 10%

Students that are 12 years or older

= 100% - (50% + 5% + 10%)

= 100% - 65%

= 35%

This means that 35% of the students are 12 years or older and we've been given the number as 70.

Let's say the total number of students is x. Therefore,

35% of x = 70

0.35 × x = 70

0.35x = 70

x = 70/0.35

x = 200

The total number of students is 200.

Therefore, the number of students that are 10years will be:

= 1/20 × 200

= 10 students

Therefore, 10 students are 10 years.

3 0
3 years ago
To close a tab, click the ____ button in the web page thumbnail on the tab switcher.
slavikrds [6]
Tabs are rectangular boxes found on top of your we browser. At the right of each of those tabs, you see a symbol written in 'x' (as shown in the picture). This is the one you should click to close an existing tab.

3 0
3 years ago
What does the Communications Decency Act of 1996 provide?
Softa [21]

Answer:

a

Explanation:

its the correct becayse

3 0
3 years ago
Which of the following is NOT a common slide layout?
Kitty [74]
The answer is going to be story, for your question of the following that is not a common slide layout
3 0
3 years ago
PLZ HELP NEEDS TO BE ANSWERED ASAP THANK YOU
Stells [14]

In python:

lst = ([])

largest = 0

while len(lst) != 6:

   user_number = int(input("Enter a number: "))

   lst.append(user_number)

   for i in lst:

       if i > largest:

           largest = i

   print(largest)

I hope this helps

7 0
3 years ago
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