Answer:
A. -0.93
Step-by-step explanation:
It is a negative due to the scatter plot line decreasing as the precipitation increases, which cancels out C & D. Then, it is decreasing rapidly, which portrays it as closer to -1 than 0, meaning the answer is A.
Part 2 is 12.5 cm represents 5km
I worked this out by doing a ratio
5cm=2km
?cm=5km
you have to do 2 times 2.5 to get 5km so you do 5 times 2.5 to get 12.5cm
so the answer is that 12.5cm
Replace x with π/2 - x to get the equivalent integral

but the integrand is even, so this is really just

Substitute x = 1/2 arccot(u/2), which transforms the integral to

There are lots of ways to compute this. What I did was to consider the complex contour integral

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

and it follows that

He would have to sell $20,000 worth of clothes to get $2,000