Question

The function n(t)=100e^-0.023t models the number of grams in a sample of cesium-137 that

remain after t years. What is the sample’s average rate of decay over the interval [3,8]?

Answer 1

Explanation:

`n(t)=100 e^(-.023t)`

when t = 3

`n(3)=100 e^(-.023* 3)`

`n(3)=100 e^(-.069)`

= 93.33

when t = 8

`n(8)=100 e^(-.023* 8)`

`n(8)=100 e^(-.184)`

= 83 .2

no of grams decayed = 93.33 - 83.2

= 10.13 grams

average rate of decay = 10.13 / ( 8 - 3 )

= 2.026 gm / year .