Question

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Answer 1

Using proportions, it is found that there would be 500,000 mites.

• This question is solved by proportions, using a rule of three.

• Each dust mite has a length of $2.5 \times 10^{-4}$ m. How many dust mites could there be in a line of $1.25 \times 10^2$ m long?

The rule of three is:

1 dust mite - $2.5 \times 10^{-4}$ m.

x dust mites - $1.25 \times 10^2$ m

Applying cross multiplication:

$2.5 \times 10^{-4}x = 1.25 \times 10^2$

$x = \frac{1.25 \times 10^2}{2.5 \times 10^{-4}}$

$x = 500000$

There would be 500,000 mites.

A similar problem is given at brainly.com/question/24372153