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# Data Sufficiency: Plugging into Yes/No Data Sufficiency

Is $$2x > x^2$$? >(1) $$0 ≤ x^2 ≤ 9$$ >(2) $$x$$ is an odd prime integer.
Incorrect. [[snippet]] Stat.(2): $$x$$ is 3 or a greater prime number (the only smaller prime is 2, which is even). This also means that $$x$$ is positive so you can divide the inequality by $$x$$ without changing the sign: >$$2x > x^2$$ >$$\displaystyle \frac{2x}{\color{red}{x}} > \frac{x^2}{\color{red}{x}}$$ >$$2 > x$$ This is __never__ true since $$x ≥ 3$$. This is a clear-cut "No," so **Stat.(2) → No → S → BD**.
Incorrect. [[snippet]] Stat.(1): If you plug in $$x=1$$, then $$0 ≤ x^2 ≤ 9$$ and $$2x > x^2$$ (since $$2 >1^2$$). But is this true for any number? No! If you plug in $$x=0$$, then $$0 ≤ x^2 ≤ 9$$ and $$2x = x^2$$ (since $$0=0^2$$). No definite answer, so **Stat.(1) → Maybe → IS → BCE**.
Incorrect. [[snippet]] Stat.(2): $$x$$ is 3 or a greater prime number (the only smaller prime is 2, which is even). This also means that $$x$$ is positive so you can divide the inequality by $$x$$ without changing the sign: >$$2x > x^2$$ >$$\displaystyle \frac{2x}{\color{red}{x}} > \frac{x^2}{\color{red}{x}}$$ >$$2 > x$$ This is __never__ true since $$x ≥ 3$$. This is a clear-cut "No," so **Stat.(2) → No → S → BD**.
Correct. [[snippet]] Stat.(1): If you plug in $$x=1$$, then $$0 ≤ x^2 ≤ 9$$ and $$2x > x^2$$ (since $$2 >1^2$$). But is this true for any number? No! If you plug in $$x=0$$, then $$0 ≤ x^2 ≤ 9$$ and $$2x = x^2$$ (since $$0=0^2$$). No definite answer, so **Stat.(1) → Maybe → IS → BCE**. Stat.(2): $$x$$ is 3 or a greater prime number (the only smaller prime is 2, which is even). This also means that $$x$$ is positive so you can divide the inequality by $$x$$ without changing the sign: >$$2x > x^2$$ >$$\displaystyle \frac{2x}{\color{red}{x}} > \frac{x^2}{\color{red}{x}}$$ >$$2 > x$$ This is __never__ true since $$x ≥ 3$$. This is a clear-cut "No," so **Stat.(2) → No → S → B**.
Incorrect. [[snippet]] Stat.(1): If you plug in $$x=1$$, then $$0 ≤ x^2 ≤ 9$$ and $$2x > x^2$$ (since $$2 >1^2$$). But is this true for any number? No! If you plug in $$x=0$$, then $$0 ≤ x^2 ≤ 9$$ and $$2x = x^2$$ (since $$0=0^2$$). No definite answer, so **Stat.(1) → Maybe → IS → BCE**.
Statement (1) ALONE is sufficient, but Statement (2) alone is not sufficient to answer the question asked.
Statement (2) ALONE is sufficient, but Statement (1) alone is not sufficient to answer the question asked.
BOTH Statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.