*20 points* What is the probability of drawing yellow marble followed by a red marble from a bag containing 12 yellow marbles, 14 red marbles, and 15 green marbles if the first marble is not replaced?

a. 192/1,849
b. 18/43
c. 21/205

Step-by-step explanation:

12 +16 +15 = 43 total marbles

12 are yellow so first pick you have 12/43 probability of picking yellow

without replacement there would be 42 marbles left

there are 16 red ones so you  have a 16/42, reduces to 8/21 probability of picking red

probability of picking yellow then red =

12/43 x 8/21 = 96/903, reduces to 32/301 probability

c: 21/205

Step-by-step explanation:

The probability of choosing a yellow marble first is 12/41 bc there are 12 yellow marbles and 41 marbles to choose from.

The probability of choosing a red marble is 14/40 bc there are 14 red marbles and 40 marbles to choose from( since you have removed the marble you first chose so there are 40 marbles left).

Multiplying these two together, 12/41 * 14/40 = 168/1640, simplified it's 21/205.

hello from mrbilldoesmath!

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

discussion:

in general, the factorization of a^3 + b^3 is

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

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

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

you,

mrb

(a + b) (a^2 - a b + b^2)

b

step-by-step explanation:

write an equation for the line of best fit. then predict the number of milk cow operations in the year 2005.