Answer:
[tex]\sqrt{48}[/tex]
Step-by-step explanation:
Looking at the numbers, we can see that the first number, third, and fourth number are all rational numbers since they either repeat or terminate.
On the other hand, let's try and simplify the square root of 48:
[tex]\sqrt{48}=\sqrt{16\cdot3}=\sqrt{16}\cdot\sqrt3=4\sqrt3[/tex]
The square root of 3 is irrational. Therefore, 4 times the square root of 3 is also irrational. All of the given numbers are rational instead of the second one. So, the second number does not belong .
