Wind (mph)Wind-Chill(∘F)52510211519201725163015351440134512501255116010 The table gives the wind chill temperature when the outside temperature is 30∘F. a. Use x as the wind speed and create a quadratic model for the data. b. At what wind speed does the model estimate that the wind chill will be 14∘F? c. Is the model found in part (a) valid for x&gt;63? Explain

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Wind (mph)Wind-Chill(∘F)52510211519201725163015351440134512501255116010  The table gives the wind chill temperature when the outside temperature is 30∘F.  a. Use x as the wind speed and create a quadratic model for the data.  b. At what wind speed does the model estimate that the wind chill will be 14∘F?  c. Is the model found in part (a) valid for x&amp;gt;63? Explain

Samantha Brown · Accepted Answer

Determining quadratic model for the data .  Below is the step-by-step procedure.  a) Let the wind speed be x.  The wind chill can be expressed as  f(x)=ax2+bx+c  from given data  y(5)=25  ⇒25=a52+b5+c  25=25a+5b+c (1)  y(10)=21  ⇒21=a102+b10+c  21=100a+10b+c (2)  y(15)=19  ⇒19=a192+19b+c  19=361a+19b+c (3)  Solving 1,2 and 3 we get  a=13315=0.043  b=−149105=−1.4190  c=195763=31.0635  wind chill y(x)=0.0413x2−1.419⊗+31.0635  b) At x=?,y(x)=14  ⇒14=0.0413x2−1.4190x+32.0635  ⇒x≈34.86.  estimate of using speed is 34.86 mph.  c) for x&amp;gt;63.  The model will be valid because it gives a general expression to determine f(x) any x.

sukljama2 · Accepted Answer

a) y=a+bx+cx2 is quadratic trend line   By using regression equation calculation required function  y=26.205−0.5014x+0.0041x2  y^=26.205−0.5014x+0.0041x2  b) Now y^=14,x=?  Putting in equation $50  14=26.205−0.5014x+0.0041x2  ⇒0.0041x2−0.5014x+12.205=0  x=0.5014±(−0.5014)2−4(0.0041)(12.205)2×0.0041  x=0.5014±0.051239960.0082  =0.5014±0.022640.0082  =88.75$33.54  c) Accepted 33.54mph  for x≥63,y=26.205−0.5214(63)+0.0041(63)2  y=42.4779−31.5882  =10.8897

alenahelenash · Accepted Answer

a) The equation of the quadratic model is, f(x)=−0.0033x2−0.4299x+31.727 ≈−0.003x2−0.429x+31.727 b) Consider the following model: f(x)=−0.003x2−0.429x+31.727. The objective is to estimate the wind speed when wind chill temperature is 21 degrees F. Plug f(x)=21 into f(x)=−0.003x2−0.429x+31.727. 21=−0.003x2−0.429x+31.727 −0.003x2−0.429x+10.727=0 This is a quadratic equation. Compare the equation with its original form ax2+bx+c=0, then a=−0.003,b=−0.429,c=10.727. Plug a=−0.003,b=−0.429,c=10.727 into the quadratic formula x=−b±b2−4ac2a x=−(−0.429)±(−0.429)2−4(−0.003)(10.727)2(−0.003) =0.429±0.312765−0.006 =0.429+0.559254−0.006 or 0.429−0.559254−0.006 =−164.709 or 21.709 Neglect the negative sign. Therefore, the wind speed when wind chill temperature 21 degrees F is 21.709mph. c) The objective is to this model is valid for x&amp;gt;72 and explain. This model is invalid because the function maximum is at x=53.7514. Thus, the model predicts warmer temperature when the wind speed above 72 mph.

\[\begin{array}{|c|c|}\hline \text{Wind (mph)}&\text{Wind-Chill} (^{\circ}F)\\ \hline 5&25\\ \hline

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