1. **State the problem:** We have the function $g(x) = \frac{1}{11}(6 \cdot 2^x + 7)$ and want to find the greatest constant $k$ such that $g(x) > k$ for all $x$. 2. **Analyze the function:** Since $2^x > 0$ for all real $x$, the term $6 \cdot 2^x$ is always positive. 3. **Find the minimum value of $g(x)$:** The smallest value of $6 \cdot 2^x$ occurs as $x \to -\infty$, where $2^x \to 0$. So, $$\lim_{x \to -\infty} g(x) = \frac{1}{11}(6 \cdot 0 + 7) = \frac{7}{11}.$$ 4. **Interpretation:** Since $g(x)$ is always greater than or equal to $\frac{7}{11}$ and approaches this value but never goes below it, the greatest constant $k$ such that $g(x) > k$ for all $x$ is $\frac{7}{11}$. **Final answer:** $\boxed{\frac{7}{11}}$