there are 12 men on the basketball team. 2 are centers, 5 are guards, the rest a forwards. what is the probability, as a percent, that out of two players chosen at random they would be a guard and a forward? ASAP PLS

the y will be in still but your answer will be y x 3

step-by-step explanation:

jk: )

$205,000 here you go i had to guess it but i got it correct so yeah

simplifying

7ln(x + 2.8) = 11.2

reorder the terms:

7ln(2.8 + x) = 11.2

(2.8 * 7ln + x * 7ln) = 11.2

(19.6ln + 7lnx) = 11.2

solving

19.6ln + 7lnx = 11.2

solving for variable 'l'.

move all terms containing l to the left, all other terms to the right.

reorder the terms:

-11.2 + 19.6ln + 7lnx = 11.2 + -11.2

combine like terms: 11.2 + -11.2 = 0.0

-11.2 + 19.6ln + 7lnx = 0.0

the solution to this equation could not be determined.

step-by-step explanation:

i searched it on google but it says that the solution to this equation could not be determined but i hope that it you a little.