Answer:

The straight line equation is $y=a_0+a_1 x$

| Year $x$ | 1951 | 1961 | 1971 | 1981 | 1991 | =9885 | | --- | --- | --- | --- | --- | --- | --- | | Production $y$ | 10 | 12 | 8 | 10 | 15 | =55 | | $xy$ | 19510 | 23532 | 15768 | 19810 | 29865 | =108485 | | $x^2$ | 3806401 | 3845521 | 3884841 | 3924361 | 3964081 | =19425205 | Normal equations are: $\sum y=na_0+a_1 \sum x$ $\sum xy=a_0 \sum x+a_1 \sum x^2$ Thus, $55=5 a_0+9885 a_1$ and $108485=9885 a_0+19425205 a_1$. Solving these two equations, we get $a_0=\dfrac{79759}{11744}, a_1=\dfrac{25}{11744}.$ Hence, a straight line is $y=a_0+a_1x=-\dfrac{79759}{11744}+\dfrac{25}{11744}.$ Answer