All multiples of 10 end in a zero besides the number zero. So 10, 20, 30 and so on....
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.
recall that a cube has all equal sides, check the picture below.
![\bf \textit{volume of a cube}\\\\ V=x^3~~ \begin{cases} x=side's~length\\[-0.5em] \hrulefill\\ V=5.12 \end{cases}\implies 5.12=x^3 \\\\\\ \sqrt[3]{5.12}=x\implies 1.72354775\approx x](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cube%7D%5C%5C%5C%5C%0AV%3Dx%5E3~~%0A%5Cbegin%7Bcases%7D%0Ax%3Dside%27s~length%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0AV%3D5.12%0A%5Cend%7Bcases%7D%5Cimplies%205.12%3Dx%5E3%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B5.12%7D%3Dx%5Cimplies%201.72354775%5Capprox%20x)
Answer:
H, 20%
Step-by-step explanation:
960/4,800 = 20/100
