Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b) 
And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:

And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10
(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>
Answer:
Point A
Explanation:
The work done by stretching or compressing a spring is given by E=1/2kx²
The potential energy is numerically equal to the work done.
This means that the higher the bigger the value of the extension, x, the higher the energy contained.
In this scenario the modulus of x is considered.
Among the given values of x the modulus of -5 is the largest.
thus it gives the highest value of energy.