Answer:
The correct answer is that a possible value for n could be all numbers from 101 to 120.
Step-by-step explanation:
Ok, to solve this problem:
You have that: [tex]10 <\sqrt{n} <11[/tex]
Then, applying the properties of inequations, the power is raised by 2 on both sides of the inequation:
[tex](10)^{2} <(\sqrt{n} )^{2} <(11)^{2}[/tex]
[tex]100<n<121[/tex]
Then, a possible value for n could be all numbers from 101 to 120.
