The table describes the cooling of a cup of coffee as it sits on your teacher’s desk in the math office. Time (min) $0, 4, 8, 12, 16, 20$ Temperature (celsius) $55, 47, 40, 34, 29, 25$
a) Calculate $a$, the cooling factor (rate at which the coffee cools per minute) of the coffee. Round your answer to two decimal places.
b) Determine an equation for $y$, the temperature of the coffee in degrees Celsius after $t$ minutes.
c) Is your model a good predictor of the temperature of the coffee at any point in time? Explain.
I had decided to solve this problem using $y=i(1-a)^t$ to solve this problem, where $y$ is the final amount, $i$ is the initial amount, $a$ is the rate of decrease or "cooling factor" in this case, and $t$ is the number of time periods. you will see why I used this formula in a little bit.
a) To solve for $a$, we can input some values from the table to fill in the variables for $i$, $y$ and $t$:
$y=47,$ $i=55,$ $t=4,$
$$47=55(1-a)^4.$$
Now from this, we can solve for $a$ to find the cooling factor:
$$\frac{47}{55}=(1-a)^4$$
$$\sqrt[4]{\frac{47}{55}}=1-a$$ $$-a=\sqrt[4]{\frac{47}{55}}-1$$ $$\sqrt[4]{\frac{47}{55}}\approx0.961$$ $$0.961-1=-0.039$$
So from this, $-a=-0.039$. We must transfer the negative sign and we get $a=0.039$. But since we are rounding to the "second decimal place, we get $\boxed{a=0.04}$.
b) Since we have found the value for $a$, We can develop our equation:
$$\boxed{y=55(1-0.04)^t=55(0.96)^t}.$$
c) From what I have provided you with, I think you will be able to answer this question.