Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (gabylovesyou):

An individual's phone number contains seven digits, not including the area code, from the set A shown below. A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Set B represents the digits in Brent's phone number. B = {5, 5, 5, 3, 0, 9, 9} Set C represents the digits in Charlie's phone number. C = {8, 6, 7, 5, 3, 0, 9} How many even numbers are in the set ∼(B ∩ C)?

OpenStudy (gabylovesyou):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

Can you form the set B ∩ C for me?

jimthompson5910 (jim_thompson5910):

any ideas?

jimthompson5910 (jim_thompson5910):

If you're not sure, then B ∩ C is the set of numbers that are in BOTH set B AND set C

OpenStudy (gabylovesyou):

would it be just 1 ? since they r EVEN numbers ?

jimthompson5910 (jim_thompson5910):

are you able to form the set B ∩ C at all?

OpenStudy (gabylovesyou):

well is it 5, 3, 0, 9 ?

jimthompson5910 (jim_thompson5910):

yep, B ∩ C = {5,3,0,9} now use this to form ~(B ∩ C)

jimthompson5910 (jim_thompson5910):

do you know what I mean?

OpenStudy (gabylovesyou):

not really lol

jimthompson5910 (jim_thompson5910):

well you basically start with set A then you erase any number you find in the set {5,3,0,9} what are you left with when you do that?

OpenStudy (gabylovesyou):

like i erase the numbers that are in set 5,3,0,9 from set A ?

jimthompson5910 (jim_thompson5910):

exactly

jimthompson5910 (jim_thompson5910):

it's like you're forming a new set by starting with all of set A...but kicking out anything you find in {5,3,0,9}

OpenStudy (gabylovesyou):

so it would be 0,1,2,4,6,7,8

jimthompson5910 (jim_thompson5910):

close, it's actually {1, 2, 4, 6, 7, 8}...you forgot to kick out 0 That's the set ~(B ∩ C) so ~(B ∩ C) = {1, 2, 4, 6, 7, 8} since there are 3 even numbers in this set (2, 4, 8) this means the answer is 3

OpenStudy (gabylovesyou):

ohhh ok :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!