For $m$ miles where $m > 1$, the cost is $3.50 + 2.25(m-1)$. The inequality is $3.50 + 2.25(m-1) \leq 20$. Expanding: $3.50 + 2.25m - 2.25 \leq 20$, so $1.25 + 2.25m \leq 20$, giving $2.25m \leq 18.75$, thus $m \leq 8.33$. Since we need whole miles, the maximum is $8$ miles. Checking: $8$ miles costs $3.50 + 2.25(7) = $19.25 \leq $20$. For $9$ miles: $3.50 + 2.25(8) = $21.50 > $20$, which exceeds the budget.

If Chen sends $$t$$ messages where $$t > 200$$, his bill is $$40 + 0.15(t - 200)$$. The inequality is $$40 + 0.15(t - 200) \leq 55$$, giving $$0.15(t - 200) \leq 15$$, so $$t - 200 \leq 100$$, thus $$t \leq 300$$. Choice B ($$250$$) is unnecessarily restrictive. Choice C ($$200$$) ignores that he can go over the base amount. Choice D ($$350$$) would cost $$40 + 0.15(150) = \62.50$$, exceeding his budget.

The temperature after $$h$$ hours is $$82 - 1.5h$$. For safety, we need $$82 - 1.5h < 78$$, which gives $$-1.5h < -4$$, so $$h > \frac{4}{1.5} = \frac{8}{3} = 2\frac{2}{3}$$ hours. Choice A ($$2.5$$ hours) is too short - the pool would still be at $$78.25°F$$. Choice B ($$3$$ hours) works but is not the minimum threshold. Choice D ($$4$$ hours) is unnecessarily long.

Setting up the inequality: $0.7(78) + 0.3F \geq 82$. This gives $54.6 + 0.3F \geq 82$, so $0.3F \geq 27.4$, thus $F \geq 91.33$. Since exam scores must be whole numbers, the minimum needed is $92$ points. Checking: with $F = 92$, the final grade is $0.7(78) + 0.3(92) = 54.6 + 27.6 = 82.2 \geq 82$. With $F = 91$, the final grade would be $54.6 + 27.3 = 81.9 < 82$, which is insufficient.

Marcus needs $$\200$$ for the laptop plus $$\25$$ for bus fare, totaling $$\225$$. He starts with $$\85$$, so he needs $$225 - 85 = \140$$ more. Setting up the inequality: $$85 + 12h \geq 225$$, which gives $$12h \geq 140$$, so $$h \geq 11.67$$. Since he can't work partial hours, he needs at least $$12$$ hours. Choice B ($$10$$ hours) gives only $$\205$$ total, not enough for both laptop and bus fare. Choice C ($$15$$ hours) is more than necessary. Choice D ($$9$$ hours) gives only $$\193$$ total.

Since Maria travels more than $5$ miles, the cost is $12 + 3(d - 5)$ where $d > 5$. The inequality is $12 + 3(d - 5) \leq 30$, giving $3(d - 5) \leq 18$, so $d - 5 \leq 6$, thus $d \leq 11$. Since we need whole miles and checking our options: For $d = 9$, cost is $12 + 3(9-5) = 12 + 12 = $24 \leq $30$. For $d = 10$, cost would be $12 + 3(5) = $27 \leq $30$. However, among the given choices, $9$ miles is the maximum option and is within budget.

Membership cost for $$n$$ movies: $$25 + 8n$$. Regular cost: $$13n$$. We want $$25 + 8n < 13n$$, which gives $$25 < 5n$$, so $$n > 5$$. This means after $$6$$ movies, membership becomes cheaper. Let's verify: At $$5$$ movies, membership costs $$\65$$, regular costs $$\65$$ (equal). At $$6$$ movies, membership costs $$\73$$, regular costs $$\78$$ (membership cheaper). Choices A, C, and D give incorrect breakeven points.

This question tests solving two-step linear inequalities from word problems, graphing solution sets on number lines (open/closed dots, directional shading), and interpreting in context, especially with whole numbers. Solving is similar to equations but preserves inequality direction—for px+q≤r: subtract q (px≤r-q), divide by p (x≤(r-q)/p if p>0, flip to x≥... if p<0); graphing uses closed dot ● for ≤ or ≥ (includes boundary), open ○ for < or > (excludes), shade left for < or ≤ (toward smaller), right for > or ≥ (toward larger); example: 5v+10≤100 → 5v≤90 → v≤18 graphs as closed dot at 18, shaded left. For needing more than 75 points with 68 current and +2 per quiz, set up 68+2q>75, solve by subtracting 68 (2q>7), dividing by 2 (q>3.5), and since quizzes are whole, q≥4 as q=4 gives 76>75 but q=3 gives 74≤75. The correct choice A reflects this with q>3.5 so q≥4 whole quizzes. Errors include using ≥75 leading to q≥3 incorrectly (choice B), dividing wrong to q>7 (choice C), or ignoring whole numbers by suggesting q=3.5 (choice D). Steps are translating 'more than' to >75, solving with inverse operations keeping direction, interpreting for whole q≥4 since context requires integers (can't do half quizzes, round up). Mistakes involve direction errors like using ≥ instead of >, forgetting to round up for strict inequality, or calculation slips like dividing 7 by 2 incorrectly.

This question tests solving two-step linear inequalities from word problems, graphing solution sets on number lines (open/closed dots, directional shading), and interpreting in context with whole numbers. Solving is similar to equations but preserves inequality direction—for px+q≤r: subtract q (px≤r-q), divide by p (x≤(r-q)/p if p>0, flip to x≥... if p<0); graphing uses closed dot ● for ≤ or ≥ (includes boundary), open ○ for < or > (excludes), shade left for < or ≤ (toward smaller), right for > or ≥ (toward larger); example: 5v+10≤100 → 5v≤90 → v≤18 graphs as closed dot at 18, shaded left for 18 or less. For this quiz problem needing more than 75 points from 68 plus 2 per quiz, set up 68+2q>75, solve by subtracting 68 (2q>7), dividing by 2 (q>3.5 since 2>0, no flip), and interpret smallest whole number as 4 quizzes since q must be integer. The correct choice A shows q>3.5 and at least 4 quizzes. Errors include using ≥ instead of > (choice C with q≥3.5), rounding down incorrectly (choice B with at least 3), or misstating as exactly 4 (choice D). Steps include: (1) translate 'more than 75' to >75, (2) solve with inverse operations, (3) for strict >, use open circle if graphing, (4) interpret with ceiling to next integer for 'at least'. Context requires whole quizzes, and mistakes involve boundary type or forgetting to round up for strict inequality.

This question tests solving two-step linear inequalities from word problems, graphing solution sets on number lines (open/closed dots, directional shading), and interpreting in context. Solving is similar to equations but preserves inequality direction—for px+q≤r: subtract q (px≤r-q), divide by p (x≤(r-q)/p if p>0, flip to x≥... if p<0); graphing uses closed dot ● for ≤ or ≥ (includes boundary), open ○ for < or > (excludes), shade left for < or ≤ (toward smaller), right for > or ≥ (toward larger); example: 5v+10≤100 → 5v≤90 → v≤18 graphs as closed dot at 18, shaded left for 18 or less. For this phone plan starting at $50 credit minus $2.50 per data day, at least $20 remaining, set up 50-2.5d≥20, solve by subtracting 50 (-2.5d≥-30), dividing by -2.5 (d≤12, flip since negative), and interpret as 12 or fewer days. The correct choice A shows d≤12 and interpretation of 12 days or fewer. Errors include no flip (choice B with d≥12), wrong constants (choices C and D with d≤30 or d≥30). Steps include: (1) translate 'at least $20' to ≥20, (2) solve by subtracting and dividing with flip for negative, (3) graph would be closed dot at 12 shaded left, (4) interpret as d≤12. Mistakes often involve forgetting to flip when dividing by negative or direction errors.