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
The answer should be B - lasts longer.
The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
Objects can have the same mass (but different <span>compositions). Only mass or volume cannot tell you if the object is solid or vo</span>lumes) or same volume (but different masses)
