Accuracy, bounds and compound measures · Extension · 精度、界限与复合量 · 拓展
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| lower bound/ˈləʊə baʊnd/ | 下界 | xià jiè |
Is the printed measurement exact?
- A rectangular panel is labelled 8.0 cm by 5.0 cm, each to the nearest 0.1 cm. Its true area is not fixed at 40 cm².
- This lesson studies lower bound 下界: The smallest possible value consistent with a stated rounding rule.
印刷的测量值是否精确？
- 一个矩形面板标注为 8.0 cm 乘 5.0 cm，均四舍五入到最接近的 0.1 cm。其真实面积并非固定为 40 cm²。
- 本课学习下界：与给定舍入规则一致的最小可能值。
Choose the mathematical structure
- A value rounded to the nearest unit u lies from stated value-u/2 up to, but usually not including, stated value+u/2. For positive quantities, combine extremes according to the operation.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
选择数学结构
- 四舍五入到最接近单位 u 的值位于 stated value - u/2 到（通常不包括）stated value + u/2 之间。对于正数量，按运算组合极值。
- 计算前请先明确允许的输入项和单位。方程应表达关系本身，而不仅仅是记录计算器按键过程。
Which description correctly defines lower bound? · 哪个描述正确定义了“下界”？
The smallest possible value consistent with a stated rounding rule. · 符合给定舍入规则的最小可能值。
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
The lengths satisfy 7.95≤L<8.05 and 4.95≤W<5.05. Since A=LW, 39.3525≤A<40.6525. For speed d/t, the largest speed uses the largest distance and smallest positive time.
通过验证案例进行推导
- 将结果与初始量进行比对。代入原始关系式，或在适当情况下对比图表与数值答案。
长度满足 7.95≤L<8.05 和 4.95≤W<5.05。由于 A=LW，故 39.3525≤A<40.6525。对于速度 d/t，最大速度使用最大距离和最小正时间。
Accuracy, bounds and compound measures · 精度、界限与复合测量
A value rounded to the nearest unit u lies from stated value-u/2 up to, but usually not including, stated value+u/2 · 四舍五入到最接近单位u的值位于陈述值-u/2起，直至但不包括陈述值+u/2的范围内。
Explain the different endpoint inclusions for a measurement rounded to 8.0 cm. · 解释四舍五入到8.0厘米时不同端点的包含情况。
Find the lower bound of 3.4 rounded to the nearest 0.1. · 求3.4四舍五入到最接近的0.1时的下界。
Half the rounding unit is 0.05. Lower bound=3.4-0.05=3.35. · 舍入单位的一半是0.05。下界=3.4-0.05=3.35。
Test a tempting shortcut
- An upper bound is not automatically achieved. Dividing upper distance by upper time does not give the largest speed. Keep enough digits in intermediate calculations.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The greatest value of positive d/t uses the greatest d and greatest t. This claim is false. Explain which definition or assumption it violates.
检验一个诱人的捷径
- 上界并非自动达成。用最大距离除以最大时间并不能得到最大速度。中间计算应保留足够位数。
- 当捷径失效时，找出其违背的假设。保留精确值直到题目要求的最终舍入步骤。
正数 d/t 的最大值使用最大的 d 和最大的 t。此说法错误。请说明它违反了哪一定义或假设。
Find the upper bound of 250 rounded to the nearest 10. · 求250四舍五入到最接近的10时的上界。
Half of 10 is 5. Upper bound=250+5=255; the upper endpoint is excluded. · 10的一半是5。上界=250+5=255；上端点不包含在内。
The greatest value of positive d/t uses the greatest d and greatest t. · 正数d/t的最大值使用最大的d和最大的t。
An upper bound is not automatically achieved. Dividing upper distance by upper time does not give the largest speed. Keep enough digits in intermediate calculations. · 上界并不自动被达到。用上限距离除以上限时间并不能得到最大速度。中间计算需保留足够位数。
Interpret a new situation
- Distinguish measurement uncertainty from arithmetic rounding. A sensible reported precision cannot be finer than the measurements justify.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
解读新情境
- 区分测量不确定性与算术舍入。合理的报告精度不能比测量所允许的更精细。
- 完整解答需给出数学结果并解释其含义。检查其在所述背景下是否可行。
A distance is at most 105 m and time at least 20 s. Find the upper speed bound. · 距离至多为105米，时间至少为20秒。求最大速度界限。
For positive d and t, the largest quotient uses the greatest d and least t: 105/20=5.25. · 对于正数d和t，最大的商使用最大的d和最小的t：105/20=5.25。
Match each part of a complete solution to its purpose. · 将完整解答的每个部分与其目的相匹配。
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · 假设用于论证模型的合理性；检查用于验证结果；解释用于将其与问题建立联系。
Use this in your course
- 9260 · Extension · 3.1. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The smallest possible value consistent with a stated rounding rule. Choose the relationship, show the method, check its assumptions and interpret the result.
将其应用于你的课程
- 9260 · 拓展内容 · 3.1。在分配拓展内容前，请先匹配目标层级和课程大纲要求。
- 在给出最终答案前先展示解题方法，并遵循试卷的计算器及公式使用规则。通过定位第一个无效步骤来复盘错误答案。
与给定舍入规则一致的最小可能值。选择关系式，展示方法，检查假设并解释结果。