One friend’s phone willindependently drop calls with 20% probability. Your other friend’s phone will independentlydrop calls with 30% probability. Say you need to make 6 phone calls, so you randomly chooseone of the two phones and you will use thatsamephone to make all your calls (but you don’tknow which has a 20% versus 30% chance of dropping calls). Of the first 3 (out of 6) callsyou make, one of them is dropped. What is the conditional expected number of dropped callsin the 6 total calls you make (conditioned on having already had

its 4 because 2 times 2 is 4, if you double the number two you get four

idk man good luck man

in if you ask questions, the points are taken from your total and given to the one who answered it..

so if i have 30 points and i ask a question for 5 points, then when someone asks a question i will lose 5 points and the answerer earns 5

happy to

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