Universal set \(\sigma\) = {7, 8, 9, 10, 11, 12, 13, 14, 15} A = {numbers that are multiples of 3} B = {odd numbers} C = {10 \(\le\) number \(\le\) 13} how do I know how many numbers there are in set \(A \cup B\)
AUB = { 7,9,11,12,13,15}
???
A={9,12,15} B = {7,9,11,13,15}
n(A)= 3 n(B)= 5 n(A intersection B) = 2 (9 and 15) so n(A union B) = n(A)+n(B)- n(A intersection B) = 3+5-2 = 6
u got that lg ?
less math more words?
hartnn's way is a analatical thing.
it is more easy to draw a diagram..
aren't there infinite numbers in set A U B?
i think this can be more difficult for you to solve as a way of hartuun
like i said before...less math more words....i really am not a ver good math learner
in case you haven't noticed my "i hate math" [picture
the universal set here is finite so how come union ever become infinite ?
oh
the universal set limits the elements
you could have said that earlier...
you said.. universal set is cofined in several number//
i thought you noticed..
yup, we see w.r.t universal set mentioned/given.
i also said i didn't know what those meant
so "numbers multiple of three" are the ones in the universal set?
universal set means that total group is that number.
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