Question

thought provoking the table shows the temperatures y (in degrees fahrenheit) of a cup of tea after x minutes. write a function that models the data and can be used to predict the temperature of the tea after 20 minutes. time, x 0 2 4 6 8 10 temperature, y 150 164 146 131 120 111 y = prediction: °f explain your reasoning.

Explanation:

Step1: Assume the function is quadratic \(y = ax^{2}+bx + c\)

Substitute \((x = 0,y = 150)\) into \(y=ax^{2}+bx + c\), we get \(c = 150\).
Substitute \((x = 2,y = 164)\) into \(y=ax^{2}+bx + 150\), we have \(4a + 2b+150=164\), which simplifies to \(2a + b=7\) (Equation 1).
Substitute \((x = 4,y = 146)\) into \(y=ax^{2}+bx + 150\), we get \(16a+4b + 150=146\), which simplifies to \(4a + b=-1\) (Equation 2).

Step2: Solve the system of equations

Subtract Equation 1 from Equation 2: \((4a + b)-(2a + b)=-1 - 7\).
\(4a + b-2a - b=-8\), \(2a=-8\), \(a=-4\).
Substitute \(a = - 4\) into Equation 1: \(2\times(-4)+b=7\), \(b = 15\).
So the function is \(y=-4x^{2}+15x + 150\).

Step3: Predict the temperature at \(x = 20\)

Substitute \(x = 20\) into \(y=-4x^{2}+15x + 150\).
\(y=-4\times20^{2}+15\times20 + 150\).
\(y=-4\times400+300 + 150\).
\(y=-1600+300 + 150=-1150\) (This is wrong, let's try exponential - decay model \(y = ab^{x}+c\))

Let's assume the function is of the form \(y = ab^{x}+c\).
When \(x = 0\), \(y=150\), so \(a + c=150\).
When \(x = 2\), \(y = 164\), \(ab^{2}+c=164\).
When \(x = 4\), \(y = 146\), \(ab^{4}+c=146\).
From \(a + c=150\), \(c = 150 - a\).
Substitute \(c = 150 - a\) into \(ab^{2}+c=164\) and \(ab^{4}+c=146\).
\(ab^{2}+150 - a=164\), \(a(b^{2}-1)=14\) (Equation A).
\(ab^{4}+150 - a=146\), \(a(b^{4}-1)=-4\) (Equation B).
Let \(u = b^{2}\), then Equation A: \(a(u - 1)=14\), Equation B: \(a(u^{2}-1)=-4\).
Since \(u^{2}-1=(u - 1)(u + 1)\), \(\frac{a(u^{2}-1)}{a(u - 1)}=\frac{-4}{14}\), \(u + 1=-\frac{2}{7}\) (not valid).

Let's use linear - regression formula.
The formula for the regression line \(y=mx + b\), where \(m=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}\), \(n = 6\), \(\sum_{i=1}^{6}x_{i}=0 + 2+4 + 6+8 + 10=30\), \(\sum_{i=1}^{6}y_{i}=150+164 + 146+131+120+111=822\), \(\sum_{i=1}^{6}x_{i}y_{i}=0\times150+2\times164+4\times146+6\times131+8\times120+10\times111\)
\(=328+584+786+960+1110=3768\), \(\sum_{i=1}^{6}x_{i}^{2}=0^{2}+2^{2}+4^{2}+6^{2}+8^{2}+10^{2}=4 + 16+36+64+100=220\)
\(m=\frac{6\times3768-30\times822}{6\times220 - 30^{2}}=\frac{22608-24660}{1320 - 900}=\frac{-2052}{420}=-4.8857\approx - 5\)
\(b=\frac{\sum_{i = 1}^{n}y_{i}-m\sum_{i = 1}^{n}x_{i}}{n}=\frac{822-(-5)\times30}{6}=\frac{822 + 150}{6}=\frac{972}{6}=162\)

The function is \(y=-5x + 162\)

Step4: Predict at \(x = 20\)

Substitute \(x = 20\) into \(y=-5x + 162\)
\(y=-5\times20+162=-100 + 162 = 62\)

Answer:

The function is \(y=-5x + 162\), and the predicted temperature after \(20\) minutes is \(62^{\circ}F\)