For the first problem, the answer is D, because every year, the graph goes down by about $4,500.
For problem two,
a. It's located in quadrant one because x and y are both positive (I've attached a graph with labeled quadrants for reference)
I'm unsure about b and c but I hope I helped with the others!
The average has to be at least 120 and at most 130
To calculate the average we need the sum of all values divided by the number of values, in this case, three (135, 145 and the third result).
120 ≤ (135 + 145 + n)/3 ≤ 130
In inequalities like this, what we change in one side, must be changed in the othe rside as well.
360 ≤ 280 + n ≤ 390
80 ≤ n ≤ 110
Yes, 104.12 is greater than 104.002
because 104.12 rounds up to 104.1
and 104.002 rounds up to 104.0
