## Related questions with answers

A strain of E.coli Beu 397-recA441 is placed into a nutrient broth at $30^{\circ}$ Celsius and allowed to grow. The data shown in the table are collected. The population is measured in grams and the time in hours. Since population P depends on time t and each input corresponds to exactly one output, we can say that population is a function of time; so P(t) represents the population at time t.

$\begin{matrix} \text{Time (hours), t} & \text{Population (grams), P}\\ \text{0} & \text{0.09}\\ \text{2.5} & \text{0.18}\\ \text{3.5} & \text{0.26}\\ \text{4.5} & \text{0.35}\\ \text{6} & \text{0.50}\\ \end{matrix}$

Find the average rate of change of the population from 4.5 to 6 hours.

Solutions

VerifiedNote that the average rate of change is same as the slope of the line joining two points.

$\text{Average rate of change}=\dfrac{\Delta y}{\Delta x} = \dfrac{0.5-0.35}{6-4.5} = \dfrac{0.15}{1.5} = 0.1\text{ grams/hour}$

This means that the population increased at an average rate of 0.1 grams/hour.

The average rate of change: $\dfrac{f(6) - f(4.5)}{6-4.5 }$

$= \dfrac{0.5 - 0.35}{1.5 } =0.1$

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Precalculus Enhanced with Graphing Utilities

6th Edition•ISBN: 9780321795465 (2 more)Sullivan## More related questions

- prealgebra

1/4

- prealgebra

1/7