Answer:
a. 63 °F
b. When i decided to model the situation, I assumed that the temperature varied inversely as the elevation and that the change in elevation or temperature was linear.
Step-by-step explanation:
a. To model this situation, we assume the temperature varies inversely as elevation decreases since at elevation 6288 ft the temperature is 56 °F and at elevation 2041 ft, the temperature is 87 °F
So, we model this as a straight line.
Let m be the gradient of the line.
Let the (6288 ft, 56 F) represent a point on the line and (2041 ft, 87 °F) represent another point on the line.
So m = (6288 ft - 2041 ft)/(56 °F - 87 °F) = 4247 ft/-31 °F = -137 ft/°F
At elevation 5376 ft, let the temperature be T and (5376 ft, T) represent another point on the line.
Since it is a straight line, any of the other two points matched with this point should also give our gradient. Since in the gradient, we took the point (6288 ft, 56 °F) first, we will also take it first in this instant.
So m = -137 ft/ °F = (6288 ft - 5376 ft)/(56 °F - T)
-137 ft/°F = 912 ft/(56 °F - T)
(56 °F - T)/912 ft = -1/(137 ft/ °F)
56 °F - T = -912 ft/(137 ft/°F)
56 °F - T = 6.66 °F
T = 56 °F + 6.66 °F
T = 62.66 °F
T ≅ 62.7 °F
T ≅ 63 °F to the nearest degree
b. When i decided to model the situation, I assumed that the temperature varied inversely as the elevation and that the change in elevation or temperature was linear.
