Acceleration = ( final velocity - intial velocity ) / time
= (60-0)m/s / 3s
= 20 m/s2
the answer is a
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
Answer:
solar eclipse
Explanation:
because at that time, the moon completely covers the path and casts its shadow on earth because it is present between sun and earth's path. so, solar eclipse occurs.
hope it helps :)
Answer:
i) 0.9504
ii) 0.0452
Explanation:
Given data: reliability of hydraulic brakes= 0.96
reliability of mechanical brakes = 0.99
So the probability of stopping the truck = 0.96×0.99= 0.9504
At low speed
case: A works and B does not
= 0.96×(1-0.99) = 0.0096
case2 : B works and A does not
= 0.99×(1-0.96) = 0.0396
Therefore, probality of stopping = 0.0096+0.0396 = 0.0492