# Question #56c3d

Jun 6, 2017

$t = \frac{k s}{e}$

#### Explanation:

This will be a variant on $y = k x$

The one thing that will be in both is the need for a constant of variation $k$

What is happening? as the number of exits increases it should be the case that the people exit faster. Thus $t$ will reduce

Let the total time of completed exist be $t$
Let the count of exits be $e$
Let the count of people (for this consideration) be the constant $s$

Then as $E$ increases we need $t$ to reduce

Logically we would have $t \to \frac{s}{e}$

However this would need to incorporate some way of allowing for adjustment. This is where $k$ comes in. Write as:

$t = k \times \frac{s}{e} \text{ "->" } t = \frac{k s}{e}$
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Note that the choice of the variant $e$ is not one I would have picked. In mathematics $e$ is a recognised constant that is reserved.

It is used a lot in log to base $e \text{ "->log_e" "->" } \ln$