Two boxes labeled A and B are full of tickets numbered either 00 or 11 for classifying and counting. (The sum of the draws equals the number of 1's in the draws and the average of the draws equals the proportion of 1's in the draws.) We know that in Box A, 36% of the tickets are labeled 11 . We also know that in Box B, 10% of the tickets are labeled 11 . We will draw at random with replacement 400 times from Box A. Independently, we will draw at random with replacement 400 times from Box B. From Box A: the sum of the 400 draws expected value equal:

choice b is correct

step-by-step explanation:

we have been given the following system of linear inequalities;

we are required to graphically determine the solution to these system of inequalities. using a graphing tool, the solution is as shown in the attachment below;

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb