Answer:
Probability ball: white - P(W)P(W);
Probability ball: even - P(E)P(E);
Probability ball: white and even - P(W&E).
Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).
(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)−0P(WorE)=P(W)+P(E)−0. Not sufficient
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)−P(E)=0.2P(W)−P(E)=0.2, multiple values are possible for P(W)P(W) and P(E)P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)P(WorE).
(1)+(2) P(W&E)=0 and P(W)−P(E)=0.2P(W)−P(E)=0.2 --> P(WorE)=2P(E)+0.2P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4P(E)=0.4 (10 even balls) then P(WorE)=1P(WorE)=1 BUT if P(E)=0.2P(E)=0.2 (5 even balls) then P(WorE)=0.6P(WorE)=0.6. Not sufficient.
Answer: E.
A=1/2x Base x Height
= 1/2 (10)(24)
=120
2 ways: Easy and hard
Hard=A
Easy=B
A: 1/2x+4
work from there so we do fun stuff with it
make something that can be simplified so
1/2x+4 times (2/2)=x+8
now square the whole thing and put the result in a square root thingie
(x+8)^2=x^2+16x+64

multiply the whole thing by 4/4 and put
![\sqrt{16} [\tex] on top so then [tex] \sqrt{x^2+16x+64}](https://tex.z-dn.net/?f=%20%5Csqrt%7B16%7D%20%5B%5Ctex%5D%20on%20top%20so%20then%20%0A%5Btex%5D%20%5Csqrt%7Bx%5E2%2B16x%2B64%7D%20)
times

=

=

to solve it, factor out the 16 in the square root and then square root 16 to get 4
then it will be (4 times square root of equation)/4=square root of equatio
factor square root of equation and square root it and get x+8
divide by 2 to get 1/2x+4
B: 1/2x+4
put stuff that cancels out
1/2x+3x-3x+4+56-56
move them around
3 and 1/2x-3x+60-56
or
2x-3x+1 and 1/2x+30-20+30-36
then just add like terms to solve