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2 years ago

A length is measured as 21cm correct to 2 significant figures. what is the upper bound for the length?

Mathematics

1 answer:

Leno4ka [110]2 years ago

7 0

the upper bound for the length is $21.5 cm$ .

<u>Step-by-step explanation:</u>

Lower and Upper Bounds

The lower bound is the smallest value that will round up to the approximate value.
The upper bound is the smallest value that will round up to the next approximate value.

Ex:- a mass of 70 kg, rounded to the nearest 10 kg, The upper bound is 75 kg, because 75 kg is the smallest mass that would round up to 80kg.

Here , A length is measured as 21cm correct to 2 significant figures. We need to find what is the upper bound for the length . let's find out:

As discussed above , upper bound for any number will be the smallest value in decimals which will round up to next integer value . So , for 21 :

⇒ $21.5 cm$

21.5 cm on rounding off will give 22 cm . So , the upper bound for the length is $21.5 cm$ .

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Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

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Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .