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Choice B is correct. It’s given that the minimum value of $x$ is $12$ less than $6$ times another number $n$. Therefore, the possible values of $x$ are all greater than or equal to the value of $12$ less than $6$ times $n$. The value of $6$ times $n$ is given by the expression $6n$. The value of $12$ less than $6n$ is given by the expression $6n-12$. Therefore, the possible values of $x$ are all greater than or equal to $6n-12$. This can be shown by the inequality $x\ge 6n-12$.

Choice A is incorrect. This inequality shows the possible values of $x$ if the maximum, not the minimum, value of $x$ is $12$ less than $6$ times $n$.

Choice C is incorrect. This inequality shows the possible values of $x$ if the maximum, not the minimum, value of $x$ is $6$ times $n$ less than $12$, not $12$ less than $6$ times $n$.

Choice D is incorrect. This inequality shows the possible values of $x$ if the minimum value of $x$ is $6$ times $n$ less than $12$, not $12$ less than $6$ times $n$.