Question

prove that the radius of curvature at any point (x, y) for
$\sqrt[3]{x} + \sqrt[3]{y} = \sqrt[3]{a}$
is 3(axy)^1/3​

Answer 1

Answer:

Total no of balls is 3+5+7=15 Favorable cases (not black) =10(3+7) P(not black) = favourable outcomes /total no of outcomes So here P(not black) =10/15=2/3 Therefore the probability that the ball drawn is not black is 2/3 Pls mark as brainliest..

Step-by-step explanation: