Write an equation in y=mx+b form for the linear relationship shown in the table. Wind-chill effect of 20-mile-per-hour wind Actual Temperature(t)Wind-Chill Temperature (w)45∘F25∘F25∘−3∘F5∘F−31∘F

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Write an equation in y=mx+b form for the linear relationship shown in the table.  Wind-chill effect of 20-mile-per-hour wind  Actual Temperature(t)Wind-Chill Temperature (w)45∘F25∘F25∘−3∘F5∘F−31∘F

Cassandra Ramirez · Accepted Answer

For First Question.  t= actual temp. degree Fahrenheit  w(t)= Wind chill temp. degree Fahrenheit.  So, the point is (t, w(t))  Or, (45,25) ,(25,-3) and (5,-31) all values in degree Fahrenheit.  So, Equation of line pass through from  (45 ,25) and (25 ,-3) is  Or , y−(−3)=(−3−2525−45)(x−25)  Or , y+3=(−28−20)(x−25)  Or , y+3=(1410)(x−25)  Or , y+3=1.4x−35  Or , y=1.4x−35−3  Or , y=1.4x−38.   Now , Check this equation for (5,-31) ,Put the value of x=5  So , y=(1.4)(5)−38  Or , y=7−38  Or , y=−31  Hence , this linear equation satisfied the all three points.   So , The final answer is: y=1.4x−38

Wendy Boykin · Accepted Answer

Solution:  Standard linear equation y=mx+b  In order to calculate slope, m and b we will use data from table  Equation w(t)=mt+b  when t=45,w=25  25=m×45+b  b=25−45m (i)  when t=25,w=−3  −3=m×25+b  −3=25m+25−45m [b from (i)]  −28=−2om  m=2820=1.4  m=1.4  Than b=25−45×1.4  b=−38  Plug these values w(t)=1.4t−38

alenahelenash · Accepted Answer

t= actual temp. degree Fahrenheit w(t)= (45,25) ,(25,-3) and (5,-31) all values in degree Fahrenheit. Equation of line pass through from (45 ,25) and (25 ,-3) is y−(−3)=(−3−2525−45)(x−25) y+3=(−28−20)(x−25) y+3=(1410)(x−25) y+3=1.4x−35 y=1.4x−35−3 y=1.4x−38. Now , Check this equation for (5,-31) ,Put the value of x=5 y=(1.4)(5)−38 y=7−38 y=−31 Hence , this linear equation satisfied the all three points. Answer: y=1.4x−38

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