Answer:
a) For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

d.9
b) 
a.15
c) For this case we have the sample size n = 25 and the sample variance is
, the standard error can founded with this formula:

Step-by-step explanation:
Part a
For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

d.9
Part b
From a sample we know that n=41 and SS= 600, where SS represent the sum of quares given by:

And the sample variance for this case can be calculated from this formula:

a.15
Part c
For this case we have the sample size n = 25 and the sample variance is
, the standard error can founded with this formula:

Let V = the launch velocity
Let θ = the launch angle
Let d = the horizontal distance traveled
Ignore air resistance.
The horizontal component of velocity is
u = V cos θ
The time of flight is
t = d/(V cosθ) = (d/V) secθ
Create a table of θ versus t as shown below.
θ t/(d/V)
---- ----------
45 1.4142
40 1.3054
35 1.2208
30 1.1547
The graph shows that as the launch angle decreases below 45°, the time of flight decreases.
Therefore arrow b (θ < 45°) arrives first.
Answer: Arrow b arrives first.
Answer:
the y intercept would be 4 pr (0,4)