cos 2x – cosx = 0
------------ (1)
where cos2x = 2cos^2x –
1
equation (1) becomes;
<span>2cos^2x – 1 – cosx = 0 ------------ (2)</span>
Suppose cosx = y, then
equation (2) becomes;
2y^2 – 1 – y = 0
By factorization method;
(2y + 1) x (y – 1) = 0
Where 2y + 1 = 0 or y –
1 = 0
So solving both;
y = -1/2 or 1
As y = cosx
So cosx = -1/2 or 1
The above equation is true only for x = 2/3π, 0, 4/3 π
<span>The solution were find out for interval [0,2 π),
where 2 π is not included in the interval, as the interval is not a closed on
the right side.</span>
The answer would be
X = 6 - y / 2 !
The residual value comes out to be 2.94 cm and height is 157.06 cm
<u>Explanation:</u>
The regression equation is calculated at the first step.
height = 105.08 plus 2.599 multiply foot length
At foot length = 20cm, height = 105.08 plus 2.599 multiply 20
= 157.06 cm
Residual = Actual minus predicted value = 160 minus 157.06
=2.94 cm
B) The residual standard deviation generally gives a sense of the goodness of fit of goodness of regression equation on our data. The magnitude tells us that how much will be predicted values from model will vary from actual values. the linear model is justified.