Hello! Will mark Brainiest!
x^3+y^3+z^3=k

Step-by-step explanation:

Assuming x, y, and z as well as k need to be positive integers, you can look at what integers have cubes under 100. Five cubed is already too high, so there are only 4 cubes you need to look at, and you can combine them to your heart’s content. I find only 9 values for k if all 4 are positive integers. If we allow 0 there are 14 more.

At first, allowing x, y and z to be any integer positive or negative doesn’t seem to expand your possibilities that much, but it expands them MASSIVELY. This is because once some values can be negative, x, y, and z can be HUGE. But it needs to sum to under 100? A massive positive number cubed plus a massive negative number cubed can be under 100. If negative numbers are allowed, everything under 100 can be the sum of 3 cubes EXCEPT: numbers which are 4 or 5 more than a multiple of 9 cannot be the sum of 3 cubes

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0