# B use the quadratic polynomial to estimate the

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(b) Use the quadratic polynomial to estimate the outdoor temperature at 7:30 am, to the nearest tenth of a degree. (work optional)
(c) Use the quadratic polynomial y = 0.3476 t 2 + 10.948 t 6.0778 together with algebra to estimate the time(s) of day when the outdoor temperature y was 75 degrees. .
Show algebraic work in solving. State your results clearly; report the time(s) to the nearest quarter hour.
10. (10 pts) EXPONENTIAL REGRESSION
Data: A cup of hot coffee was placed in a room maintained at a constant temperature of 69 degrees, and the coffee temperature was recorded periodically, in Table 1. TABLE 1 REMARKS: Common sense tells us that the coffee will be cooling off and its temperature will decrease and approach the ambient temperature of the room, 69 degrees. So, the temperature difference between the coffee temperature and the room temperature will decrease to 0. We will fit the temperature difference data (Table 2) to an exponential curve of the form y = Ae bt . Notice that as t gets large, y will get closer and closer to 0, which is what the temperature difference will do. So, we want to analyze the data where t = time elapsed and y = C 69, the temperature difference between the coffee temperature and the room temperature. TABLE 2 t = Time Elapsed (minutes) C = Coffee Temperature (degrees F.) 0 166.0 10 140.5 20 125.2 30 110.3 40 104.5 50 98.4 60 93.9 t = Time Elapsed (minutes) y = C 69 Temperature Difference (degrees F.) 0 97.0 10 71.5 20 56.2 30 41.3 40 35.5 50 29.4 60 24.9
0 10 20 30 40 50 60 70 0 20 40 60 80 100 120 f(x) = 89.98 exp( -0.02 x ) R² = 0.98 Temperature Difference between Coffee and Room Time Elapsed (minutes) Temperature Difference (degrees) Exponential Function of Best Fit (using the data in Table 2): y = 89.976 e 0.023 t where t = Time Elapsed (minutes) and y = Temperature Difference (in degrees)