WEBVTT
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So in this problem we are given that researchers measured
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blood alcohol concentration of eight men Uh starting one hour
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after consumption of a couple of drinks. And we're
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giving a table rooty isn't ours. One 1.5,
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2.2.5 and 3.0 and the concentration C of T in
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milligrams per millimeter 0.31 0.24, 0.18, 0.12 And
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0.08. First of all, we're asked to find
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the average change of C with respect to T over
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each time interval. So going from 1.02, we
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have 0.18-0.3, 1 Over 2-1 Is a
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-0.18. This is milligrams per millimeter and time is
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in ours. Okay. And from 1.5 two,
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We have 0.18-0.24 over 2-1.5. So that
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a negative 0.06 over 0.5. So that's a negative
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0.12 again, milligrams per mil later our Okay,
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The next one from 2.0 To 2.5. Well,
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from our table This will be 0.1, two-0.18
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. We're just using our average rate of change equation
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here between these two timeframes, keeping everything coordinated minus
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two. and so I have a negative 0.06 over
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0.5, Which again is a negative 0.12 milligrams per
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million. Later our Okay. And the last one
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here, we're going from 2.0 23 oh. And
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so that's 0.08 minus 0.18 over 3.0-2. And
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so that's a negative 0.10 over one. That's a
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negative 0.10 milligrams per million later. Power. All
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right. In part b says to estimate the instantaneous
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rate of change, estimate the instantaneous right of change
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. Ah T equals two. Okay, so then
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at T equals two. What we can do is
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come up here and think about what happens as we
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approach two from each direction left and right. Well
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from 1.5 to 2 we did right here A-
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-12. And from right here we did it from
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the right And got a negative 0.12. So we
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have the same rate coming in from both sides.
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So normally we would average these two. But since
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they're the same, that means we're going to have
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the same instantaneous rate as we have a constant rate
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coming in from both sides left and right.