Answered

The data set represents a progression of hourly temperature measurements. Use a graphing calculator to determine the quadratic regression equation for this data set.

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
[tex]$y$[/tex] & 20 & 16 & 10 & 0 & -7 & -20 \\
\hline
\end{tabular}

a. [tex]$y = 0.795x^2 + 3.796x + 20.180$[/tex]

b. [tex]$y = -0.795x^2 - 3.760x + 20.180$[/tex]

c. [tex]$y = -0.875x^2 - 3.596x + 20.179$[/tex]

d. [tex]$y = -0.795x^2 - 3.796x + 20.180$[/tex]



Answer :

GinnyAnswer
Let's analyze the given data set of hourly temperature measurements and find the quadratic regression equation that best fits these points.

The data points are:

[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 20 & 16 & 10 & 0 & -7 & -20 \\ \hline \end{array} \][/tex]

In determining the quadratic regression equation, we need to find the coefficients \(a\), \(b\), and \(c\) of the general quadratic equation:

[tex]\[ y = ax^2 + bx + c \][/tex]

The coefficients obtained are approximately:

1. \(a \approx -0.875\)
2. \(b \approx -3.596 \)
3. \(c \approx 20.179\)

Thus, the quadratic regression equation that fits the given data set is:

[tex]\[ y = -0.875 x^2 - 3.596 x + 20.179 \][/tex]

Comparing this equation with the options provided:

a. \(y = 0.795 x^2 + 3.796 x + 20.180\)
b. \(y = -0.795 x^2 - 3.760 x + 20.180\)
c. \(y = -0.875 x^2 - 3.596 x + 20.179\)
d. \(y = -0.795 x^2 - 3.796 x + 20.180\)

The correct answer is:

c. \(y = -0.875 x^2 - 3.596 x + 20.179\)

So, the quadratic regression equation that best represents the data set is:

[tex]\[ \boxed{y = -0.875 x^2 - 3.596 x + 20.179} \][/tex]

Other Questions