Ajar contains 4 red marbles numbered 1 to 4 and 10 blue marbles numbered 1 to 10. a marble is drawn at random from the jar. find the probability the marble is blue or evennumbered.\

Step-by-step explanation:

As per given ,  The jar contains

Red marbles = (R1) , (R2) , (R3) , (R4)

Blue marbles =  (B1) ,   (B2) ,   (B3) ,   (B4) ,   (B5) ,   (B6) ,   (B7) ,   (B8) ,   (B9) ,   (B10) .

Total marbles = 4+10=14

The marbles that has even number = (R2)  , (R4) ,(B2) , (B4) ,  (B6) ,   (B8) , (B10)

=7

Total Blue marbles  = 10

Blue and even marbles = 5

Now , the number of marbles are blue or even numbered :

n(Blue or even )= n(Blue) + n(even)- n(Blue and even)

= 10+7-5 =12

Now , the probability the marble is blue or even numbered will be :

Hence, required probability =

i think the answer is true

step-by-step explanation:

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)