Answer:
Ok, as i understand it:
for a point P = (x, y)
The values of x and y can be randomly chosen from the set {1, 2, ..., 10}
We want to find the probability that the point P lies on the second quadrant:
First, what type of points are located in the second quadrant?
We should have a value negative for x, and positive for y.
But in our set; {1, 2, ..., 10}, we have only positive values.
So x can not be negative, this means that the point can never be on the second quadrant.
So the probability is 0.
Answer:
Step-by-step explanation:
b/c the function goes thur the point (1,4) we know it's 4 times more than f(x) so g(x) = 4![x^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D)
Answer:(-1,2), (6,8), (5,-7)
Step-by-step explanation:
The answer is a I hope this helped you out
Answer:
1/10
Step-by-step explanation:
7/10
= .7