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anonymous
 one year ago
The volume V, in liters, of air in the lungs during a foursecond respiratory cycle is approximated by the model
V = 0.1729t + 0.1522t2 − 0.0374t3,
where t is the time in seconds. Approximate the average volume of air in the lungs during one cycle. (Round your answer to four decimal places.)
anonymous
 one year ago
The volume V, in liters, of air in the lungs during a foursecond respiratory cycle is approximated by the model V = 0.1729t + 0.1522t2 − 0.0374t3, where t is the time in seconds. Approximate the average volume of air in the lungs during one cycle. (Round your answer to four decimal places.)

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1calculate the area by integration and divide by 4 dw:1443087401475:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1which will give you a time weighted average dw:1443087484413:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the answer would be 2?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dunno let me integrate ....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1mmm, don't fancy all those decimals but the the machine should get it right this is the integral http://www.wolframalpha.com/input/?i=int_%7Bt%3D0%7D%5E%7B4%7D+%28+0.1729t+%2B+0.1522t%5E2+%E2%88%92+0.0374t%5E3%29+dt and that blue rectangle i drew was just for illustration, it does not purport to be an actual solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm getting two with a few different decimal combinations.. none are correct. ugh

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1the integral is there for you to check against..... the machine does the indefinite integral as well as the definite so you can nail it properly then as i said you are looking for a timeweighted average what's curious is that the graph goes back to zero at \(t \approx 5\) whereas you are asked for the average over \(0<t<4\) but i suppose it is just a model

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay I plugged in my intergrals & got 1.19892 but that didn't work UGH
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