Free math worksheets

- Identify every dot strictly to the right of \(12.4\) on the value axis. - Repeated masses must be included once for each dot. - Add only the measurements that satisfy the stated condition.

1. The plotted masses greater than \(12.4\,\text{g}\) are \(12.5\,\text{g}\), \(12.5\,\text{g}\), \(12.8\,\text{g}\), and \(13.0\,\text{g}\). 2. Add them: \(12.5+12.5+12.8+13.0=50.8\). 3. The total mass is \(50.8\,\text{g}\).

\(50.8\,\text{g}\)

- Mark the two endpoints of the qualifying interval on the horizontal scale. - Values exactly at an endpoint qualify because the interval is inclusive. - Every dot outside the interval must be removed; no dot inside it needs to be removed.

1. Values below \(8.8\,\text{s}\) do not qualify, so the \(8.4\,\text{s}\) and \(8.6\,\text{s}\) trials must be removed. 2. Values above \(9.5\,\text{s}\) do not qualify, so the \(9.7\,\text{s}\) trial must also be removed. 3. All other plotted times are inside the inclusive interval. 4. Therefore, the smallest possible number removed is \(3\).

Remove \(3\) trials: \(8.4\,\text{s}\), \(8.6\,\text{s}\), and \(9.7\,\text{s}\).

- Convert the occupied horizontal-axis values, not the dot counts. - The same observation stays the same observation when its unit changes. - Preserve each stack's frequency after converting its measurement value.

1. Convert each occupied liter value to milliliters: \(0.8\,\text{L}=800\,\text{mL}\), \(1.0\,\text{L}=1000\,\text{mL}\), \(1.2\,\text{L}=1200\,\text{mL}\), and \(1.5\,\text{L}=1500\,\text{mL}\). 2. Unit conversion changes the horizontal values but does not change how many observations occurred at each value. 3. The new plot would have \(1\) dot at \(800\), \(2\) dots at \(1000\), \(1\) dot at \(1200\), and \(1\) dot at \(1500\) milliliters.

\(800\,\text{mL}\): \(1\) dot \(1000\,\text{mL}\): \(2\) dots \(1200\,\text{mL}\): \(1\) dot \(1500\,\text{mL}\): \(1\) dot

- Treat dots at the same measurement in the two plots as belonging to one combined stack. - Compare the combined frequencies, not the frequencies in either plot alone. - For the range, use the least and greatest values appearing anywhere in the two sets.

1. Combining the plots gives three pieces at \(14.5\,\text{cm}^2\) and three pieces at \(15.0\,\text{cm}^2\). No other area occurs as often. 2. The least combined area is \(14.0\,\text{cm}^2\), and the greatest is \(15.5\,\text{cm}^2\). 3. The range is \(15.5-14.0=1.5\,\text{cm}^2\).

a) \(14.5\,\text{cm}^2\) and \(15.0\,\text{cm}^2\), with \(3\) pieces each b) \(1.5\,\text{cm}^2\)

- Only dots to the left of \(1.0\) need additional water. - For each such container, compare its amount with the target rather than adding the plotted amount itself. - Add the separate shortfalls after finding them.

1. The containers below \(1.0\,\text{L}\) contain \(0.7\,\text{L}\), \(0.9\,\text{L}\), and \(0.9\,\text{L}\). 2. Their shortfalls are \(0.3\,\text{L}\), \(0.1\,\text{L}\), and \(0.1\,\text{L}\). 3. Add the shortfalls: \(0.3+0.1+0.1=0.5\). 4. A total of \(0.5\,\text{L}\) must be added.

\(0.5\,\text{L}\)

- Round each plotted measurement separately before combining equal rounded values into stacks. - Pay particular attention to the value exactly halfway between two whole seconds. - Find the range from the least and greatest values after rounding, not before.

1. The times \(10.2\) and \(10.4\) round to \(10\). The times \(10.5\), \(10.6\), \(11.2\), and \(11.4\) round to \(11\). 2. The rounded plot therefore has \(2\) dots at \(10\,\text{s}\) and \(4\) dots at \(11\,\text{s}\). 3. The rounded least value is \(10\,\text{s}\), and the rounded greatest value is \(11\,\text{s}\), so the range is \(1\,\text{s}\).

a) \(2\) dots at \(10\,\text{s}\); \(4\) dots at \(11\,\text{s}\) b) \(1\,\text{s}\)

- Apply the correction to every measurement, including both extreme values. - Range depends on the difference between the greatest and least values. - Consider what happens to that difference when the same amount is added to both numbers.

1. The original least temperature is \(21.5^\circ\text{C}\), so the corrected least is \(22.0^\circ\text{C}\). 2. The original greatest temperature is \(22.3^\circ\text{C}\), so the corrected greatest is \(22.8^\circ\text{C}\). 3. The corrected range is \(22.8-22.0=0.8^\circ\text{C}\). 4. The original range is also \(22.3-21.5=0.8^\circ\text{C}\). Adding the same amount to every measurement shifts both extremes equally, so their difference does not change.

a) Least: \(22.0^\circ\text{C}\); greatest: \(22.8^\circ\text{C}\) b) \(0.8^\circ\text{C}\) c) The range does not change because the same \(0.5^\circ\text{C}\) is added to every reading.

- The pairing rule changes every measurement by the same multiplicative factor. - Apply that factor to each dot separately, including repeated measurements. - Find the new range only after identifying the new least and greatest masses.

1. Double each plotted mass: \(0.4\to0.8\), \(0.5\to1.0\), \(0.5\to1.0\), \(0.6\to1.2\), and \(0.8\to1.6\) kilograms. 2. The new masses are \(0.8, 1.0, 1.0, 1.2, 1.6\) kilograms. 3. The new range is \(1.6-0.8=0.8\,\text{kg}\).

a) \(0.8\,\text{kg}, 1.0\,\text{kg}, 1.0\,\text{kg}, 1.2\,\text{kg}, 1.6\,\text{kg}\) b) \(0.8\,\text{kg}\)

- Find the least and greatest occupied values in each plot separately. - Compute each range before comparing the batches. - Use the problem's definition of “more consistent”; no other statistic is needed.

1. Batch A ranges from \(48.5\,\text{g}\) to \(51.0\,\text{g}\), so its range is \(51.0-48.5=2.5\,\text{g}\). 2. Batch B ranges from \(49.0\,\text{g}\) to \(50.5\,\text{g}\), so its range is \(50.5-49.0=1.5\,\text{g}\). 3. Batch B has the smaller range. The difference between the ranges is \(2.5-1.5=1.0\,\text{g}\).

Batch B; its range is \(1.0\,\text{g}\) smaller.

- Put each dot into exactly one of the two groups using the stated cutoff. - Remember that “at most” includes \(1.0\), while “more than” does not. - Compare the two group totals only after summing each group separately.

1. The amounts at most \(1.0\,\text{L}\) are \(0.8\,\text{L}\), \(1.0\,\text{L}\), and \(1.0\,\text{L}\). Their total is \(0.8+1.0+1.0=2.8\,\text{L}\). 2. The amounts greater than \(1.0\,\text{L}\) are \(1.2\,\text{L}\), \(1.4\,\text{L}\), and \(1.6\,\text{L}\). Their total is \(1.2+1.4+1.6=4.2\,\text{L}\). 3. The greater-than-\(1.0\) group contains more juice by \(4.2-2.8=1.4\,\text{L}\).

At most \(1.0\,\text{L}\): \(2.8\,\text{L}\) More than \(1.0\,\text{L}\): \(4.2\,\text{L}\) The greater-than-\(1.0\) group contains \(1.4\,\text{L}\) more.

- The greatest value is fixed by the condition in the problem. - Work backward from the required range to determine the least value that would produce it. - Check that your proposed measurement is below the current least value so it actually changes the range.

1. If \(2.9\,\text{cm}\) remains the greatest value and the new range must be \(0.8\,\text{cm}\), the new least value must satisfy \(2.9-\text{least}=0.8\). 2. Therefore, the least value must be \(2.1\,\text{cm}\). 3. The current least value is \(2.4\,\text{cm}\), so the new measurement must be \(2.1\,\text{cm}\) to create that new least value.

\(2.1\,\text{cm}\)

- Identify the two smallest occupied measurements before changing the plot. - Joining without overlap means the two areas are added. - After updating the dots, identify the new least and greatest values before finding the new range.

1. The two smallest areas are \(2.0\,\text{cm}^2\) and \(2.5\,\text{cm}^2\). Their combined area is \(2.0+2.5=4.5\,\text{cm}^2\). 2. The dots at \(2.0\) and \(2.5\) disappear. One new dot is placed at \(4.5\), where there is already one dot. 3. The original range is \(5.0-2.0=3.0\,\text{cm}^2\). 4. After the change, the least area is \(3.0\,\text{cm}^2\), and the greatest remains \(5.0\,\text{cm}^2\). The new range is \(5.0-3.0=2.0\,\text{cm}^2\).

a) \(4.5\,\text{cm}^2\) b) The dots at \(2.0\) and \(2.5\) disappear; one new dot is added at \(4.5\). c) Before: \(3.0\,\text{cm}^2\); after: \(2.0\,\text{cm}^2\)