One container is filled with a mixture that is 30% acid. A second container is filled with a mixture that is 50% acid. The second container is 50% larger than the first, and the two containers are emptied into a third container. What percent of acid is the third container?

42% acid

Step-by-step explanation:

Suppose that these two storage containers are filled to complete capacity?

That makes the second container have 3 volume parts and the first container having 2 volume parts.

v*(1.50)= (3/2)*v

First Container     2v       30% acid

Second Container    3v       50% acid

The mixture of all of this is in the third container. Two parts plus three parts is 5 parts for the mix volume.

=(2*30+3*50)/5

=210/5

=42

So,

42 % Acid, mixture

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