Question

the temperature (in degrees celsius) as a function of time (in hours since noon) for a certain day in san jose, costa rica, is graphed. what is the approximate average rate at which the temperature decreases between 3:30 pm and 8:30 pm? choose 1 answer. 0.4 degrees celsius per hour 0.8 degrees celsius per hour 1.2 degrees celsius per hour 1.6 degrees celsius per hour

Explanation:

Step1: Determine the time values

3:30 pm is \(3.5\) hours since noon and 8:30 pm is \(8.5\) hours since noon. The change in time \(\Delta t=8.5 - 3.5=5\) hours.

Step2: Estimate the temperature values

From the graph, at \(t = 3.5\) (3:30 pm), \(T(3.5)\approx24\) degrees Celsius. At \(t = 8.5\) (8:30 pm), \(T(8.5)\approx18\) degrees Celsius. The change in temperature \(\Delta T=18 - 24=- 6\) degrees Celsius.

Step3: Calculate the average rate of change

The formula for the average rate of change of a function \(y = f(x)\) is \(\frac{\Delta y}{\Delta x}\). Here, for the function \(T(t)\) (temperature as a function of time), the average rate of change is \(\frac{\Delta T}{\Delta t}\). Substitute \(\Delta T=-6\) and \(\Delta t = 5\) into the formula: \(\frac{\Delta T}{\Delta t}=\frac{-6}{5}=-1.2\) degrees Celsius per hour. The magnitude (since we are asked about the rate at which it decreases) is \(1.2\) degrees Celsius per hour.

Answer:

C. \(1.2\) degrees Celsius per hour