Answer:
We want to know the temperature after 20 minutes if we replace x =20 we got:
[tex] y = 180.3 -2.5*20 = 130.3[/tex]
Now we need to find how many minutes did it take to cool from 180.3 degrees to 100.3 degrees, so we can set up the following equation:
[tex] 100.3 = 180.3 - 2.5x[/tex]
And solving for x we got:
[tex] x = \frac{100.3-180.3}{-2.5}= 32 minutes[/tex]
Step-by-step explanation:
For this case we can use a lineal model given by:
[tex] y= mx+b[/tex]
Where mis the slope and b the initial value so then based on the info provided the model would be:
[tex] y = 180.3 -2.5 x[/tex]
Where x represent the time and y the temeperature.
We want to know the temperature after 20 minutes if we replace x =20 we got:
[tex] y = 180.3 -2.5*20 = 130.3[/tex]
Now we need to find how many minutes did it take to cool from 180.3 degrees to 100.3 degrees, so we can set up the following equation:
[tex] 100.3 = 180.3 - 2.5x[/tex]
And solving for x we got:
[tex] x = \frac{100.3-180.3}{-2.5}= 32 minutes[/tex]
