NONLINEAR MODELS-For the latter part of the quis, we will explore some noanlisea
ID: 3114489 • Letter: N
Question
NONLINEAR MODELS-For the latter part of the quis, we will explore some noanlisear models. 4 (16 ps) QUADRATIC REGRESSION Data: On a particular somimer day, the outdoor temporature was recorded at S times of the day, and the fellowing table was compiled A scaterplot was produced and the parabola of best fit was determined Temperature on a Summer Day Time ys Outdoor of day Temperature 70 60 50 thour idegrees 52 67 73 76 0.3476+10.948 6.0778 0 9699 14 17 20 s0 61 12 Time of Day (hour) 16 20 Quadratic Poly nomial of Best Fit y--Q3476r. 10.94&-6.0778 where ,-Time of day (hour, and y-Temperature (in degrees) REMARKS. The times are the hours since midmight. For instance, 7 means 7 am, and 13 means 1 pm ta) Using algebeaic techniques we have learned, find the maximem temperature predicted by the quadratic model and find the time when it occurred. Report the time to the nearest quarter hour (ie, 00or-isor-30 or-45) (For nstance, a time of 18,25 hours is reportod as 6:15 pm.) Report the maximam temperature to the nearest tenth of a degree Show algebraic work (b) Use the quadratic polynomial to estimate the outdoor temperature at 7:30 am, to the ncarest tenth of a degree. (work optional) 03476r 10.948-6.0778 sogether with algebra to estimate the timets) of ic) Use the quadratic polymomial y day whes the outdoor temperature y was 72 drgrees That is, solve the quadratic oquation 72-0.3476r 10.948-6.0778 s) to the nearest quartor hourExplanation / Answer
y = -0.3476t^2 + 10.948t -6.0778
maximum temperature is 80.127 degrees at 15.748 hour ( 3 .45 pm )
b) outdoor temperature at 7.30 am is
plugging t = 7.5
y = -0.3476(7.5)^2 + 10.948(7.5) -6.0778
y = 56.5 degrees
c) 72 = -0.3476t^2 + 10.948t -6.0778
t = 10.96 and 20.56
so , at 11 am and 8:30 pm the temperature is around 72 degrees
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