State the gradient of the line parallel to 3y = 5x – 1.
1 answer:
You might be interested in
Given the equation
which models the data tabulated below:
![\begin{tabular} {|c|c|} x&y\\[1ex] 0.5&10.4\\ 1&5.8\\ 2&3.3\\ 3&2.4\\ 4&2 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7C%7D%0Ax%26y%5C%5C%5B1ex%5D%0A0.5%2610.4%5C%5C%0A1%265.8%5C%5C%0A2%263.3%5C%5C%0A3%262.4%5C%5C%0A4%262%0A%5Cend%7Btabular%7D)
The linear regression equation is given by
where:
and 
We extend the given table as follows:
![\begin{tabular} {|c|c|c|c|} x&y&x^2&xy\\[1ex] 0.5&10.4&0.25&5.2\\ 1&5.8&1&5.8\\ 2&3.3&4&6.6\\ 3&2.4&9&7.2\\ 4&2&16&8\\[1ex] \Sigma x=10.5&\Sigma y=23.9&\Sigma x^2=30.25&\Sigma xy=32.8 \end{tabular} \\ \\ \\ \bar{x}= \frac{\Sigma x}{n} = \frac{10.5}{5} =2.1 \\ \\ \bar{y}=\frac{\Sigma y}{n} = \frac{23.9}{5} =4.78](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%20%7B%7Cc%7Cc%7Cc%7Cc%7C%7D%20x%26y%26x%5E2%26xy%5C%5C%5B1ex%5D%200.5%2610.4%260.25%265.2%5C%5C%201%265.8%261%265.8%5C%5C%202%263.3%264%266.6%5C%5C%203%262.4%269%267.2%5C%5C%204%262%2616%268%5C%5C%5B1ex%5D%20%5CSigma%20x%3D10.5%26%5CSigma%20y%3D23.9%26%5CSigma%20x%5E2%3D30.25%26%5CSigma%20xy%3D32.8%20%5Cend%7Btabular%7D%20%5C%5C%20%5C%5C%20%5C%5C%20%5Cbar%7Bx%7D%3D%20%5Cfrac%7B%5CSigma%20x%7D%7Bn%7D%20%3D%20%5Cfrac%7B10.5%7D%7B5%7D%20%3D2.1%20%5C%5C%20%5C%5C%20%5Cbar%7By%7D%3D%5Cfrac%7B%5CSigma%20y%7D%7Bn%7D%20%3D%20%5Cfrac%7B23.9%7D%7B5%7D%20%3D4.78)
Thus,
and
Therefore, the linearlized form of the equation is y = 9.234 - 2.12x
Part B:
At x = 1.6,
which models the data tabulated below:
The linear regression equation is given by
where:
We extend the given table as follows:
Thus,
and
Therefore, the linearlized form of the equation is y = 9.234 - 2.12x
Part B:
At x = 1.6,