Question

The rate at which water flows out of a pipe in gallons per hour is given by a differentiable function R of time t. The table shows the rate as measured every 3 hours for a 24 hr period.
Hours 0 R(t) 9.6
Hours 3 R(t) 10.4
Hours 6 R(t) 10.8
Hours 9 R(t) 11.2
Hours 12 R(t) 11.4
Hours 15 R(t) 11.3
Hours 21 R(t) 10.2
Hours 24 R(t) 9.6

The rate at which water flows out of a pipe, can be approximated by q(t)= 1/79(768+23t-t^2) Use q(t) to approximate the average rate of the water flow during the 24 hour time period. Indicate units of measure.

Answer 1

using the equation q(t)=(1/79)[768+23t-t^2] we get the fol
Hours 0 R(t)= 9.72
Hours 3 R(t) =10.48
Hours 6 R(t) =11.01
Hours 9 R(t) =11.32
Hours 12 R(t)= 11.39
Hours 15 R(t)= 11.24
Hours 21 R(t)= 10.25
Hours 24 R(t)= 9.42 get the total here to find the average
-------------------
Rtotal=84.83
now average R=84.83/8=10.60 gal/hour ...ans

Answer 2

use your calculator in the eq q(t)=(1/79)[768+23t-t^2]

Answer 3

let me know if you have questions

Answer 4

@mark_o thank you, but I don't know how to use my graphing calculator to figure it out.