eq 2 $, cancel $ t - 2 $: $ G(t) = t + 2 $. The domain excludes $ t = 2 $, so $ t \in (-\infty, 2) \cup (2, \infty) $. Simplified form: $ G(t) = t + 2 $, domain: $ oxed{(-\infty, 2) \cup (2, \infty)} $.Question: If a linguist randomly selects three distinct letters from the English alphabet, what is the probability that they form a placeholder typographic symbol, given that there are 10 such symbols and 26 letters?

February 22, 2026

$eq 2 $, cancel $ t - 2 $: $ G(t) = t + 2 $. The domain excludes $ t = 2 $, so $ t \in (-\infty, 2) \cup (2, \infty) $. Simplified form: $ G(t) = t + 2 $, domain: $ oxed{(-\infty, 2) \cup (2, \infty)} $.Question: If a linguist randomly selects three distinct letters from the English alphabet, what is the probability that they form a placeholder typographic symbol, given that there are 10 such symbols and 26 letters?$

Solution: Solve $ 3^k - rac{3^{2k}}{2} < -1 $. Let $ x = 3^k $ (note $ x > 0 $). The inequality becomes $ x - rac{x^2}{2} < -1 $, or $ -rac{x^2}{2} + x + 1 < 0 $. Multiply by $-2$ (reversing inequality): $ x^2 - 2x - 2 > 0 $. Solve $ x^2 - 2x - 2 = 0 $: roots are $ x = 1 \pm \sqrt{3} $. Since $ x > 0 $, critical point is $ x = 1 + \sqrt{3} pprox 2.732 $. The quadratic is positive when $ x > 1 + \sqrt{3} $. Since $ x = 3^k $, find smallest $ k $ such that $ 3^k > 1 + \sqrt{3} $. Test $ k = 1
Question: A plant biologist models the growth efficiency of a drought-resistant crop with $ G(t) = rac{t^2 - 4}{t - 2} $. Simplify $ G(t) $ and determine its domain.
Solution: Factor numerator: $ t^2 - 4 = (t - 2)(t + 2) $. Thus, $ G(t) = rac{(t - 2)(t + 2)}{t - 2} $. For $ t
Solution: The total number of ways to choose 3 distinct letters is $inom{26}{3}$. The favorable cases involve selecting 3 letters that correspond to exactly one placeholder symbol. Assuming each symbol is defined by a unique combination of 3 letters (with no overlap in symbol definitions), there are 10 favorable combinations. Thus, the probability is $rac{10}{inom{26}{3}} = rac{10}{2600} = rac{1}{260}$. $oxed{\dfrac{1}{260}}$
Question: If a historian randomly arranges 5 historical documents, 2 of which are letters and 3 are manuscripts, what is the probability that the two letters are adjacent?
Solution: Treat the two letters as a single entity. This gives $4!$ arrangements for the entities (the letter pair and 3 manuscripts). The letters within the pair can be ordered in $2!$ ways. Total favorable arrangements: $4! imes 2!$. Total possible arrangements: $5!$. The probability is $rac{4! imes 2!}{5!} = rac{2}{5}$. $oxed{\dfrac{2}{5}}$

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