A histogram rectangle has a horizontal width and a vertical height. When its height represents frequency density, the count for that class is obtained from width multiplied by height. Looking only at the top of the rectangle leaves out part of the information needed to recover the frequency.

In the invented time example, the first class runs from zero up to two minutes and contains twelve observations. Its width is two, so its density is twelve divided by two, or six. The next class runs from two up to six minutes and contains twenty observations. Its width is four and its density is five.

The first bar is taller because six is greater than five. The second bar has the larger area in data-axis units because four times five is twenty, while two times six is twelve. It therefore represents the larger count. Both statements can be true without any contradiction in the graph.

A child who has worked mainly with equal-width frequency histograms may expect height to answer every count question directly. That earlier experience is useful, but the learner must inspect the new axis and widths before transferring the shortcut. The difficulty is often a reading decision rather than multiplication itself.

Parents can ask, “What does the vertical axis measure, and how wide is this class?” Let the child state those two facts before comparing counts. This makes the required operation visible. The tallest bar still tells something meaningful about concentration per unit of the horizontal variable; it simply does not, by itself, settle which unequal-width class contains the greatest number of observations. Keeping those two questions separate is the central repair developed throughout this article.

Frequency is the number of observations belonging to a stated class. If twelve recorded times lie in the interval 0 ≤ t < 2 minutes, the class frequency is twelve. It counts records, not minutes, squares on the graph or the width of the interval.

The class definition matters. A recorded time of 1.5 minutes belongs to this first class, while a time of exactly two minutes does not. In our example, two belongs to the next class, 2 ≤ t < 6. The inequalities ensure that a boundary value is assigned once rather than counted in both neighbouring groups.

A frequency of twelve says nothing exact about where the twelve observations lie inside the interval. They might cluster near zero, near two or across several different values. Grouping preserves the count while discarding some detail about the individual observations. This limitation becomes important when a later question asks about part of a class.

Parents can ask the child to finish the sentence, “There are twelve observations with times from zero minutes inclusive to two minutes exclusive.” That wording checks the count, the measured variable and the boundary convention together. It is more precise than saying simply that “the first bar is twelve”.

The fictional dataset in this article is for mathematical explanation. It does not describe actual eduKate students, lesson performance or waiting times at a Punggol location. Counts in a real question should be read from its own table or correctly interpreted graph. Once the student knows what each class includes, the next decision is how that count has been encoded visually. The number of observations stays the same whether it appears directly as a frequency height or indirectly through a frequency-density rectangle.

Class width is the upper boundary minus the lower boundary, measured in the horizontal variable’s units. For 0 ≤ t < 2, the width is 2 − 0 = 2 minutes. For 2 ≤ t < 6, it is 6 − 2 = 4 minutes. For 6 ≤ t < 10, it is 10 − 6 = 4 minutes.

The inclusion or exclusion of an endpoint does not add an extra minute to a continuous interval’s width. A child who calculates 6 − 2 + 1 may be counting integer labels rather than measuring the span of this time interval. These are different tasks and should not be mixed.

For the complete invented example, the three frequencies are twelve, twenty and eight. The table below keeps the interval, width, frequency and density in separate columns. Read across one row to connect its quantities, then compare the rows to see why the tallest density bar need not have the largest frequency.

Use the stated boundaries rather than guessing from a sketch’s physical appearance. If a horizontal axis is labelled in minutes, subtract its numerical boundary values. Counting small squares is useful only after each square’s numerical value has been established. A wide-looking bar on an unscaled drawing does not supply a class width.

Parents can ask, “Which two boundary values did you subtract?” This makes the calculation checkable. Keep the units attached to the width and use a positive width for each nonempty interval. Once class width is secure, density and frequency become a simple reversible pair of operations. The learner can divide a count by width when drawing a density histogram, or multiply density by width when reading its count.

Frequency density is the class frequency divided by class width. It describes the average count per unit of the horizontal variable within that class. In our time example, twelve observations over a two-minute interval give a density of six observations per minute of class width.

This phrase does not mean that six people arrive every minute or that a lesson produces six results per minute. It is a description of how recorded values are grouped along the time axis. The variable’s meaning must be kept distinct from a real-world flow rate that the dataset has not measured.

The second class contains twenty observations over four minutes, giving density five. It has more observations in total but fewer per minute of interval width on average. The third class contains eight over four minutes, giving density two. These heights allow classes with different widths to be displayed on a consistent area basis.

The rectangle area in data-axis units is width times density: w × (f/w) = f for positive width w. This algebra explains why the frequency is preserved in the area. The graph is not merely replacing one unfamiliar label with another; it is adjusting the height so that a wider class does not gain extra area without accounting for its count.

Parents can ask the learner to explain both operations. To find density, divide frequency by width. To recover frequency, multiply density by width. A clear explanation should use the particular class values and identify the axis units. If the child can repeat the formula but cannot say what the numerator and denominator represent, pause on the table before moving to a finished graph. Understanding the pair of quantities makes the formula easier to select appropriately.

If a histogram’s vertical axis is explicitly labelled frequency, a bar’s height gives its class count. Equal-width classes commonly use this presentation. Read the axis rather than assume that every histogram uses density or that every histogram uses raw frequency.

Consider three equal-width time classes, each spanning two minutes, with frequencies twelve, twenty and eight. A frequency-height histogram would have heights twelve, twenty and eight. The tallest bar represents twenty observations because the vertical scale is directly counting observations.

If the same equal-width data are instead plotted with frequency density, the heights are six, ten and four. They are not the raw counts, but their ordering matches the frequencies because each was divided by the same width two. Multiplying every height by two recovers the counts.

This explains why a height-comparison habit can seem successful for a long time. With equal widths, height and frequency are proportional on a density histogram. The shortcut works for comparing which count is larger, yet it does not justify reporting the height itself as the count unless the axis is frequency or the necessary width factor is one.

Parents can ask two separate questions: “Can I compare counts by heights here?” and “Can I read the count directly from the height?” Equal density-class widths allow the first, but not automatically the second. A learner should answer using the axis label and common width. Beginning with these familiar cases makes the unequal-width extension easier to understand. The goal is to preserve the correct earlier method while attaching the conditions that explain when it works, rather than replace every frequency graph with an unnecessary density calculation.

A wider class covers a larger interval of the measured variable. On a frequency-density histogram, the height describes concentration per unit width, so a wider rectangle can contain more observations even if its height is lower. Both dimensions are needed to compare total counts.

Our first density bar has width two and height six, giving twelve observations. The second has width four and height five, giving twenty. A height-only comparison would choose the first class, while an area calculation correctly chooses the second for the question about the greatest count.

Now imagine a separate example with width two and density six compared with width four and density three. Both classes have frequency twelve. Their bars have different heights but equal areas in data-axis units. Doubling the width while halving the density can leave the frequency unchanged.

A larger width does not by itself guarantee more observations either. A class of width ten with density one represents ten observations, fewer than the width-two, density-six class’s twelve. “Choose the widest bar” is therefore no better as a general count rule than “choose the tallest bar”.

Parents can ask for one multiplication beside each bar before making a comparison. This turns a visual judgment into an explicit count check. The student can then describe the result accurately: one class has higher density, another may have higher frequency, and two may have equal frequency despite different shapes. These are different mathematical relationships, all visible through the same width-height product. The practical habit is to decide what the question asks before choosing which visual feature to compare, rather than assume that the most striking rectangle must be the one with the most observations.

A small graph square is a drawing unit. Its numerical meaning comes from the axis labels. If one horizontal square represents two minutes, a bar spanning two squares has width four minutes. If one vertical square represents density 2.5, a height of two squares represents density five.

That rectangle then represents frequency four times five, or twenty. Multiplying its two-square width by its two-square height gives four square units of drawing area, not automatically four observations. The conversion from graph squares to data-axis units must be included.

In this example, one small rectangular grid cell corresponds to horizontal width two multiplied by vertical density 2.5, giving five observations of area. Four such cells therefore represent twenty observations. Counting grid cells can work, but only after this area conversion has been established.

Read several labelled ticks when the scale is unfamiliar. A horizontal axis labelled zero, four and eight over equal physical gaps tells you the numerical step; it does not prove that each small square equals one. Similarly, the density axis may use decimals or a multiplier that needs to be respected.

Parents can ask, “What numerical value does one square represent on each axis?” Then ask the child to use boundary values and the bar-top value directly. This reduces dependence on counting squares, especially when a rectangle spans fractional squares. A correct formula with incorrectly read scales still gives an incorrect count. Separating the scale reading from the area multiplication helps identify which step needs attention and gives the learner a method that works across differently sized diagrams without assuming that all classroom graph paper uses the same numerical units.

When class width and density are known, multiply them to recover the class frequency. Suppose a bar spans 10 ≤ x < 16 and has density 3.5. Its width is six, so its frequency is 6 × 3.5 = 21. The decimal height does not require a decimal count in this example.

A student might copy 3.5 as the frequency because it is the only value read vertically. Ask what the horizontal span contributes. Density has already divided the count by width, so the recovery step restores that factor rather than treating the height as a standalone answer.

If the graph describes counts and an exactly specified width-height product produces a non-integer, inspect the situation. The bar height may have been read approximately, the drawing may be imprecise, or the graph may represent an estimated or differently scaled quantity. Do not automatically round without considering the information and instructions supplied.

The equation f = wd can be written beside the calculation, with f for frequency, w for class width and d for frequency density. Make these symbols explicit rather than assume the child remembers them from another worksheet. The numerical values must all belong to the same class.

Parents can ask for a reverse check: divide the recovered twenty-one by the width six and confirm density 3.5. This checks the arithmetic pair but does not independently prove that the graph was read correctly, so keep the boundary and axis readings visible too. A useful solution contains the class width, the density and their product. That short chain is easier to inspect than a final number with no indication of how the student used the rectangle’s two dimensions.

Some problems provide one class frequency while leaving the density axis incompletely labelled. That known rectangle can establish the scale. First read its numerical class width, then divide the supplied frequency by that width to find the density represented by its top.

Suppose a reference bar has width four and frequency twenty. Its density is five. If its top lies two small vertical squares above zero, each vertical square represents density 2.5 on that linear axis. This is a calibration, not a guess based on how tall the bar looks.

A second bar of width two and height three vertical squares would then have density 7.5 and frequency fifteen. The calculation uses the shared vertical scale. It is not necessary for the reference class and target class to have equal widths, provided both widths are correctly read.

Another route compares data-area ratios. If the target rectangle’s area in the same graph coordinates is three-quarters of the reference rectangle’s area, its frequency is three-quarters of twenty, or fifteen. The common graph scaling cancels. This approach still depends on measuring both width and height rather than their tops alone.

Parents can ask which supplied fact fixes the scale. If neither a density scale nor a frequency-area calibration is available, a drawn rectangle may show relative areas without giving absolute counts. Recognising missing information is part of accurate graph reading. The learner should not invent a conversion because a question asks for a number. In a properly specified exercise, the known class gives the needed anchor, and the remaining calculations should follow from that anchor consistently. Keep the reference class and its units visible so the calibration can be checked against the original information.

Neighbouring classes need an unambiguous rule for assigning boundary observations. Our invented dataset uses 0 ≤ t < 2, 2 ≤ t < 6 and 6 ≤ t < 10. Each lower boundary is included and each upper boundary is excluded. A time of exactly two minutes belongs to the second class only.

Other questions may use a different consistent convention, such as 0 < t ≤ 2 followed by 2 < t ≤ 6. Then exactly two belongs to the first class. The rectangle widths remain two and four, but the assignment rule changes. Read the stated inequalities rather than replace them with a favourite convention.

A label such as “2–6” can be abbreviated classroom notation, but the complete question should clarify how shared endpoints are treated when that matters. A parent can help by asking what happens to an observation equal to the boundary. This makes any ambiguity visible before totals are calculated.

For continuous measured data rounded to a stated precision, class boundaries may require interpretation from the recording rule. That is a separate topic to use only where the actual task supplies it. Do not automatically add or subtract 0.5 from every interval; such adjustments depend on how the values were recorded and the class definitions used.

Parents can ask the child to place two or three boundary values into exactly one class each. This checks classification without requiring a graph. Once the classes are mutually exclusive and collectively cover the relevant recorded values, summing their frequencies counts each observation once. Correct density arithmetic cannot repair an original table that double counts its endpoints, so the boundary decision belongs at the beginning of the process rather than as an afterthought to a polished histogram.

For non-overlapping classes covering the dataset, add the class frequencies. In our example the counts are twelve, twenty and eight, so the total is forty observations. On the frequency-density histogram, those same counts come from the three width-height products.

The calculation is 2 × 6 + 4 × 5 + 4 × 2 = 12 + 20 + 8 = 40. Adding the heights gives 6 + 5 + 2 = 13, which is not the total frequency. Adding the widths gives ten minutes of overall interval coverage, which is also not a count.

Writing one product per class helps prevent the student from changing methods halfway through. A table column headed “recovered frequency” can make the total easy to check. The sum should agree with any independently supplied sample size when the graph includes all observations and no exclusions.

A zero-frequency interval contributes zero even though it may occupy horizontal space. A histogram should preserve that interval’s position rather than compress the axis to remove it. An empty span does not become a positive count merely because its width is greater than zero; its density is zero.

Parents can ask what each term in the total represents. The answer should name a class count rather than “a height”. This is a useful checkpoint before proportions, cumulative frequencies or estimated means, all of which need the right weights. If the total is wrong, later calculations may be arithmetically tidy yet describe the wrong dataset. Establish the counts first, then reuse them consistently. The graph and frequency table should be two representations of the same forty observations, not two separate lists of numbers that the child handles with unrelated procedures.

A class proportion is its frequency divided by the total number of observations. In our forty-observation example, the first class has proportion 12/40 = 0.3, the second 20/40 = 0.5 and the third 8/40 = 0.2. These proportions add to one.

Expressed as percentages, the classes contain thirty, fifty and twenty per cent of the observations. The second class has the greatest proportion because it has the greatest frequency. Its frequency density is still lower than the first class’s density, since its interval is twice as wide.

Density and proportion answer different questions. Density compares average concentration per unit of interval width. Proportion tells what fraction of the entire dataset lies in a particular class. Neither quantity should be read as the other merely because both may be shown on a vertical axis.

For a density histogram, use area to obtain the class frequency before dividing by the total. Here the first rectangle’s area is twelve, so its proportion is twelve divided by forty. Dividing its height six by the sum of heights thirteen would give a different number with no valid class-proportion interpretation for these unequal widths.

Parents can ask, “What is the denominator in this comparison?” The total count belongs beneath a class count when finding a proportion. The class width belongs beneath a class count when finding frequency density. Keeping those denominators separate repairs a common confusion between two kinds of division. A clear answer names its quantity and, where relevant, its units. This becomes especially useful when a question presents a graph in percentages or relative density, where familiar-looking numbers do not automatically mean raw observations.

The grouped frequency tells us the count in the complete interval, not the exact distribution inside it. The class 2 ≤ t < 6 contains twenty observations, but the table alone does not tell us exactly how many lie in 2 ≤ t < 4. They could be unevenly distributed across the wider interval.

If a task asks for an estimate using a uniform-within-class model, half the interval width receives half the class frequency. The two-minute subinterval from two up to four is half the four-minute class, so its estimated count is ten. That is an estimate based on an additional assumption, not a recovered exact fact.

The constant height of the histogram rectangle represents an average density across the class. It does not prove that the original recorded values were evenly spread. A drawing can support a convenient model while still concealing individual observations that would be required for an exact count.

Under the same explicitly stated estimation assumption, the interval 0 ≤ t < 4 would contain the first class’s exact count twelve plus an estimated ten from the second class, giving an estimated twenty-two. The status of the final result follows the uncertain component even though part of the calculation is exact.

Parents can ask, “Does the boundary cut through a class, or does it line up with a class boundary?” Aligned complete classes permit exact counting from the grouped table. A cut inside a class usually needs more data or an estimation assumption. This distinction helps the child avoid presenting a confident exact answer simply because a rectangle can be divided neatly on paper. Follow the actual question’s wording and state the model when using proportional area for a partial interval.

An estimated mean from grouped data uses class midpoints as representative values and class frequencies as weights. For our intervals, the midpoints are one, four and eight minutes. The corresponding frequencies are twelve, twenty and eight.

The weighted total is 1 × 12 + 4 × 20 + 8 × 8 = 12 + 80 + 64 = 156. Divide by the total frequency forty to obtain an estimated mean of 3.9 minutes. The estimate does not assert that every observation actually equals its class midpoint.

A child who uses density heights as weights would calculate with six, five and two instead of twelve, twenty and eight. For unequal widths, that gives the wrong relative representation of the class counts. Recover frequency first by multiplying each density by its width, then use those counts in the grouped-mean calculation.

The midpoint itself is the average of the two boundaries. For 2 ≤ t < 6, it is (2 + 6)/2 = 4. It is not the class width, which happens to be four here. In the first class, the midpoint is one while the width is two; this separates two quantities that can coincide accidentally in another row.

Parents can ask the child to label a table column “midpoint” and another “frequency” before multiplying. Each product represents a midpoint contribution repeated for the number of observations in that class. The exact mean would require the original values, unless additional information establishes the relevant within-class totals. A grouped histogram can support a useful estimate, but the solution should preserve that word. The task is to calculate from the information available while recognising the detail that grouping has removed.

A relative-density histogram uses class proportion divided by class width as its height. Equivalently, its height is frequency divided by total frequency and then divided by width. Its bar areas represent proportions, so their total area in data-axis units is one.

For the forty-observation example, the three heights are 12/(40 × 2) = 0.15, 20/(40 × 4) = 0.125 and 8/(40 × 4) = 0.05. Multiplying by widths gives areas 0.3, 0.5 and 0.2. These match the class proportions and add to one.

This is an extension for learners whose current material uses relative density. The ordinary frequency-density heights in earlier chapters were six, five and two, whose areas sum to forty. The graph’s shape is proportionally rescaled, but its vertical-axis meaning and the quantity represented by area have changed.

If the total sample size forty is known, multiply a relative bar area by forty to recover a count. For the second class, width four times height 0.125 gives proportion 0.5, and 0.5 × 40 gives twenty observations. Without a supplied total, a proportion graph does not determine the absolute count.

Parents can ask, “Does area represent a count or a proportion on this graph?” The answer comes from the axis definition and normalisation. This prevents a student from applying frequency = width × height to every density-like label without checking the scaling. The basic rectangle principle remains useful, but the resulting area must be named correctly. NIST’s histogram reference distinguishes count, relative-count and area-normalised displays; the linked background reading is for graph conventions, not a claim about which version your child’s school requires.

Background on graph normalisation: NIST/SEMATECH e-Handbook — Histogram. This reference explains display conventions; it does not specify a Singapore school syllabus.

Choosing different class boundaries can change the histogram’s appearance even when the underlying observations are unchanged. A wider class combines information that narrower classes might display separately. It is therefore useful to distinguish features of the recorded data from features produced partly by the chosen grouping.

In our example, the first two classes together cover 0 ≤ t < 6 and contain thirty-two observations. If they are merged into one class, its width is six and its density is 32/6, or sixteen-thirds. The merged bar no longer shows the original difference between densities six and five.

The count thirty-two is preserved, but detail about its allocation between the two original intervals is lost in the merged representation. If only the merged count is supplied, it cannot be separated back into twelve and twenty exactly without additional information.

This explains why comparing two histograms needs care. Different class widths, boundaries or vertical-axis definitions may make bars look different for reasons beyond a change in the observations. Read the graph construction before concluding that one group is more concentrated or more variable from a casual glance.

Parents can ask what information is retained by a merge and what is removed. The total count across the combined interval remains known, while the smaller-class counts are no longer recoverable from the merged bar alone. Avoid inventing equal splits to redraw finer classes unless the task explicitly permits an estimate. A histogram is a useful summary rather than a complete record of every value. Understanding that role helps the learner explain its limits calmly, while still using the reliable frequencies and intervals that the graph genuinely provides.

Start with an equal-width frequency histogram and ask the student to read a class count from the axis. Then present equal-width frequency-density bars and ask what multiplication recovers the count. This separates axis reading from the later unequal-width comparison.

Next use the two contrasting classes from the opening: width two with density six and width four with density five. Ask which has higher density and which has higher frequency. The learner should identify the first for density and the second for count, explaining both without treating the results as contradictory.

Observe where the method breaks. An incorrect class width calls for boundary and scale work. Correct widths with height copied as frequency call for an area relationship repair. Correct counts with a wrong total or weighted mean may indicate a later arithmetic or data-summary issue instead.

In an eduKatePunggol small group of three, students can compare their readings and explain the width-height products. One can identify the interval, another state the density and another recover the count, followed by individual practice so that each learner makes all three decisions independently.

Parents considering Secondary 4 Mathematics tuition in Punggol can bring a graph their child has misread and ask which quantity they mistook for which. A useful lesson target is, “Read the axes, calculate class width and recover frequency before comparing unequal-width bars.” Ask for a changed graph with a different scale as a checkpoint. Where frequency density is beyond the learner’s current material, keep the immediate work on equal widths and accurate axis interpretation. Diagnosis should fit the task the child is actually studying rather than turn every graph difficulty into a reason to introduce more advanced content prematurely.

Ask your child to read both axis labels aloud before looking for the tallest bar. This small pause identifies the measured variable and whether the vertical numbers are frequencies, densities, proportions or another defined quantity. It gives the calculation a purpose before the child chooses an operation.

Then ask for the lower and upper boundaries of one class. Let the learner subtract them and state the width in the variable’s units. If a square-count answer appears, ask how many numerical units each square represents rather than supply the final width immediately.

For a frequency-density graph, ask, “What count does this width-height product represent?” The child can calculate it and explain why the result belongs to that particular interval. Use a second class with a different width to check whether the same relationship is applied consistently.

Avoid adding unrelated questions when the learner is still clarifying the axis. A short comparison between two bars may be enough to establish the immediate repair. Estimated means and partial-class estimates can wait until the frequency reading is secure and those tasks are relevant to the current worksheet.

Finish with a new pair of widths and densities that the child handles independently. Record the specific uncertain step if confusion remains and bring it to the teacher or tutor. Parents can provide a clear observation without completing every multiplication or teaching an entire statistics unit. Progress is visible when the learner notices the width before making a count claim, can distinguish higher density from higher frequency and knows when grouped data support only an estimate. That explanation is more informative than a correct number copied from the example just discussed.

Practice A gives 0 ≤ t < 2 with frequency twelve. The width is two and frequency density is six. Practice B gives 2 ≤ t < 6 with frequency twenty. Its width is four and density five. The second class has more observations even though its density bar is shorter.

Practice C adds 6 ≤ t < 10 with frequency eight. Its width is four and density two. The total count is twelve plus twenty plus eight, or forty. The three class proportions are 0.3, 0.5 and 0.2, obtained from counts rather than from the sum of density heights.

Practice D gives a bar covering 10 ≤ x < 16 with density 3.5. Width six times density 3.5 gives frequency twenty-one. Practice E gives width two and density 7.5, so its frequency is fifteen. Check each by dividing the frequency back by the width.

Practice F asks for the estimated count in 2 ≤ t < 4 using the uniform-within-class assumption for Practice B. Half the interval width gives an estimated ten. The grouped count alone does not establish that exact number. Practice G uses midpoints one, four and eight with frequencies twelve, twenty and eight to give estimated mean 156/40 = 3.9.

Practice H asks which class in the original example has the highest density and which has the greatest frequency. The first is highest in density; the second is greatest in frequency. State both intervals, not just bar numbers. Complete the questions before checking these explanations, and follow any accuracy or estimation instruction if the examples are adapted into a school-style exercise. Correct labels and a reason for the chosen operation are part of the practice, alongside the arithmetic.

First identify the measured variable and read the vertical-axis label completely. Determine whether the displayed height is a frequency, frequency density, proportion or relative density. A familiar rectangle shape does not fix its numerical meaning, so this reading decision comes before selecting a formula.

Read the class boundaries and calculate width from their difference in numerical axis units. Respect the stated endpoint convention and check that the relevant classes do not overlap. If graph squares are used, convert both horizontal and vertical scales before multiplying them.

On a frequency-density histogram, recover each needed frequency by multiplying width by density. Compare those counts when the task asks which class contains more observations. Compare density heights when it asks about concentration per unit width. Keep the two answers separate, especially when the class widths differ.

Use recovered counts consistently for totals, proportions and grouped-mean weights. If a question cuts through a class, decide whether the original grouped data are enough for an exact answer or whether a stated estimation model is required. Preserve the word “estimated” where the calculation depends on within-class assumptions or midpoint representatives.

For parents, the useful checkpoint is a changed graph that your child can explain independently. They should be able to say why a shorter, wider density bar can contain more observations and show the products that establish it. The Secondary 4 programme and linked statistics articles below provide further routes. Start with the unclear quantity, repair the reading step and let the next example test that understanding. This keeps revision specific and manageable, with the graph serving as a clear representation of the data rather than a collection of heights to copy without checking what they measure.