3 ms·
x = 1, p(x) -> 1 p(x) = 0.3x + 0.7x² p(x) = 0.5x + 0.5x²
by bufferoverflow 8y ago
x = 1, p(x) -> 1
p(x) = 0.3x + 0.7x²
p(x) = 0.5x + 0.5x²
- ColinWright 8y agoCoefficients have to be non-negative integers.
- bufferoverflow 8y agoYour text doesn't mention integers.
- ColinWright 8y agoDamn. That might be a typo - stand by. (pause) You were right, that was a typo, I've edited it to insert the missing word. Thank you for finding that error. I've also changed it to say that x has to be an integer.