Answer:
to find range of interquartile sets of data
you see 5 vertical lines 1 is start point 2 is start of 2nd quartile 3 is start of 3rd quartile and also represents the median. the 4th start of 4th quartile and 5th is very end point.
Therefore the interquartile is from start line 2 to start line 4
apply to each by if it says 10 and 34 then we show workjings 34-10 = 24 and that is the range once we deduct data from each other.
Therefore full range is start line 1 to end line 5
if this says 2 to 45 then full range is 45-2 = 43 and apply to each boxplot.
Step-by-step explanation:
Answer:
a = 1565217.39 ft / s ^ 2
t = 0.001725 seconds
Step-by-step explanation:
The first thing is to use the same system of units therefore we will pass the 28 inches to feet, like this:
28 in * (1 ft / 12 in) = 2.33 ft
Now yes, we can continue, we have the following data:
vi = 0
vf = 2700 ft / s
the equations in this case are as follows:
vf = vi + a * t
vf = a * t
rearranging for a
a = vf / t (1)
now with the position equation we know that:
x = vi * t + (a * t ^ 2) / 2
x = (a * t ^ 2) / 2 (2)
now replacing (1) in (2), we are left with:
x = (vf / t) * (t ^ 2) / 2
knowing that x would be 2.33 ft, which is when the cannonball exits the cannon.
2.33 = 2700 * t / 2
t = 2.33 * 2/2700 = 0.001725 seconds.
and now replace in (1)
a = vf / t = 2700 / 0.001725 = 1565217.39 ft / s ^ 2
I don’t know the 8 percent of what you do
