Y 03476t 2 10253t 04771 r² 09699 0 10 20 30 40 50 60

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y = -0.3476t2+ 10.253t -0.4771R² = 0.96990102030405060708004812162024Temperature (degrees)Time of Day (hour)Temperatureon a Spring Day
(c)Use the quadratic polynomialy=-0.3476t2+ 10.253t-0.4771together with algebra toestimate the time(s) ofday when the outdoor temperatureywas 62 degrees.That is, solve the quadratic equation62 =-0.3476t2+ 10.253t-0.4771Show algebraic work in solving.State your results clearly; report the time(s) to the nearest quarter hour..
10. (10 pts)+(extra credit at the end)EXPONENTIAL REGRESSION[Section 6.5]Data:A cup of hot coffee was placed in a room maintained at a constant temperature of 69 degrees, and thecoffee temperature was recorded periodically, in Table 1.TABLE 1t= TimeElapsed(minutes)C= CoffeeTemperature(degrees F.)0166.010140.520125.230110.340104.55098.46093.9REMARKS:Common sense tells us that the coffee will be cooling offand its temperature will decrease and approach the ambienttemperature of the room, 69 degrees.So, the temperature difference between the coffeetemperature and the room temperature will decrease to 0.We will fit the temperature difference data (Table 2) to anexponential curve of the formy=Ae-bt.Notice that astgets large,ywill get closer and closer to 0,which is what the temperature difference will do.So, we want to analyze the data wheret= time elapsedandy=C-69, the temperaturedifferencebetween thecoffee temperature and the room temperature.TABLE 2t= TimeElapsed(minutes)y=C-69TemperatureDifference(degrees F.)097.01071.52056.23041.34035.55029.46024.9
Exponential Function of Best Fit (using the data in Table 2):y= 89.976e-0.023twheret= Time Elapsed (minutes) andy= Temperature Difference (in degrees)(a) Use the exponential function to estimate the temperature differenceywhen 70 minutes have elapsed. Reportyour estimated temperature difference to the nearest tenth of a degree.(explanation/work optional)
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Term
Fall
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Math, Algebra, Completing The Square, Quadratic equation

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