Rounding to the nearest centimetre puts the true length within half a centimetre of 79 cm, and half of 1 cm is 0.5 cm.

Lower end: \(79 - 0.5 = 78.5\). A length of exactly 78.5 cm does round to 79 cm, so it is a possible length and the interval is closed at that end: \(78.5 \leq x\).

Upper end: \(79 + 0.5 = 79.5\). A length of exactly 79.5 cm would round up to 80 cm instead, so the interval is open at that end: \(x < 79.5\).

\(78.5 \leq x < 79.5\)

Half of 100 is 50, so the true number is no more than 50 below 1700:

\(1700 - 50 = 1650\)

A crowd of 1650 does round to 1700, so 1650 is itself possible — this end of the interval is included.

At the top end, 1750 would round to 1800 rather than 1700, so the true number is below 1750 — that end of the interval is not included.

People are counted in whole numbers, so the greatest possible number is the whole number just below 1750:

\(1750 - 1 = 1749\)

One decimal place means the capacity has been rounded to the nearest 0.1 litres, and half of 0.1 is 0.05.

Lower end: \(5.8 - 0.05 = 5.75\). A capacity of exactly 5.75 litres rounds up to 5.8 litres, so it is possible and the interval is closed at that end: \(5.75 \leq x\).

Upper end: \(5.8 + 0.05 = 5.85\). A capacity of exactly 5.85 litres would round to 5.9 litres instead, so the interval is open at that end: \(x < 5.85\).

\(5.75 \leq x < 5.85\)

Rounding to the nearest centimetre puts the true length within 0.5 cm of 54 cm.

The smallest it can be is \(54 - 0.5 = 53.5\), and 53.5 cm does round to 54 cm, so that end is included.

The largest it can be is \(54 + 0.5 = 54.5\), but 54.5 cm would round up to 55 cm, so that end is not included.

\(53.5 \leq x < 54.5\)

Truncating throws away everything after the whole number of centimetres rather than rounding it. A length of 54.5 cm truncates to 54 cm even though it would round up to 55 cm.

Every length from 54 cm up to 55 cm truncates to 54 cm. A length of exactly 54 cm truncates to 54, so the bottom end is included; a length of exactly 55 cm truncates to 55, not 54, so the top end is not.

\(54 \leq x < 55\)

Notice the difference from part (a): the truncated interval sits entirely above 54, while the rounded interval sat half below 54 and half above it.

Rounding to the nearest metre puts the length within half a metre of 4 m.

Lower end: \(4 - 0.5 = 3.5\), possible, so \(3.5 \leq l\).

Upper end: \(4 + 0.5 = 4.5\), not possible, so \(l < 4.5\).

\(3.5 \leq l < 4.5\)

The first plank could be anywhere from 3.5 m up to but not including 4.5 m.

The second plank could be anywhere from 10.5 m up to but not including 11.5 m.

The stated lengths give \(4 + 11 = 15\) m, but Joel's total is 14 m, which is smaller than that, so both lengths must be taken towards the lower end of their intervals.

To round to 14 m, the total must be at least \(13.5\) m and under \(14.5\) m.

Take the first length as \(3.7\) m and the second as \(10.6\) m. Each is inside its interval.

\(3.7 + 10.6 = 14.3\) m

This total is inside the range for 14 m, so it is 14 m to the nearest metre, and Joel could be correct.

To the nearest 10 cm, the true length is within half of 10 cm, that is 5 cm, of 290 cm.

Lower end: \(290 - 5 = 285\), which is possible, so the interval is closed at that end.

Upper end: \(290 + 5 = 295\), which would round up, so the interval is open at that end.

\(285 \leq x < 295\)

The total is largest when every garden trellis is as long as it can be, so use the upper bound of one garden trellis.

The upper bound is half of 10 cm above 110 cm: \(110 + 5 = 115\) cm, which is \(1.15\) m.

\(2 \times 1.15 = 2.3\) m

Each garden trellis is less than 1.15 m, so the total is less than 2.3 m. Since \(2.3 < 2.31\), the total is less than 2.31 m, so Amara is correct.

The interval \(89.5 \leq y < 90.5\) is the one for rounding, because it sits half a step either side of 90.

Truncation only discards digits after the units digit, so it never rounds up. A value such as 89.5 truncates to a number smaller than 90.

Truncating to a whole number cuts a value back to 90 only if it lies between 90 and the next step up, which is 91.

90 itself truncates to 90, so the lower end is included and takes \(\leq\). 91 truncates to 91, not 90, so the upper end is excluded and takes \(<\).

The error interval is therefore \(90 \leq y < 91\) (compare the interval for rounding, which would run half a step either side of 90).

The first two significant figures of 3100 are the 3 and the 1, and the 1 sits in the hundreds column. So 3100 has been rounded to the nearest 100, and half of 100 is 50.

Lower end: \(3100 - 50 = 3050\). 3050 rounds up to 3100, so it is possible and the interval is closed at that end: \(3050 \leq n\).

Upper end: \(3100 + 50 = 3150\). 3150 rounds up to 3200 rather than 3100, so the interval is open at that end: \(n < 3150\).

\(3050 \leq n < 3150\)

Only the first digits are given here, with more digits cut off after them — that is truncation, not rounding to the nearest 0.1.

Every value from 3.1 up to (but not including) 3.2 begins with those same digits, because anything beyond 3.1 only changes a digit further along than the ones shown or written down.

A value of exactly 3.1 does begin with 3.1, so the lower end is included: \(3.1 \leq y\).

A value of 3.2 would begin with 3.2 instead, so the upper end is not included: \(y < 3.2\).

Combining both ends gives the error interval \(3.1 \leq y < 3.2\).

(If this had instead been rounded to 1 decimal place, the true value could also lie below 3.1 — truncation only ever cuts digits away, so the interval sits entirely at or above the shown value.)

Rounding to the nearest centimetre puts the true length within half a centimetre of 64 cm, and half of 1 cm is 0.5 cm.

Lower end: \(64 - 0.5 = 63.5\). A length of exactly 63.5 cm does round to 64 cm, so it is a possible length and the interval is closed at that end: \(63.5 \leq x\).

Upper end: \(64 + 0.5 = 64.5\). A length of exactly 64.5 cm would round up to 65 cm instead, so the interval is open at that end: \(x < 64.5\).

\(63.5 \leq x < 64.5\)