Question

PLEASE HELP FAST!!

Dogs in the GoodDog Obedience School win a blue ribbon for learning how to sit, a green ribbon for learning how to roll over, and a white ribbon for learning how to stay. There are 100 dogs in the school. 73 have blue ribbons, 39 have green ribbons, and 62 have white ribbons. 21 have a blue ribbon and a green ribbon; 28 have a green ribbon and a white ribbon; 41 have a blue ribbon and a white ribbon. 14 have all three ribbons.

b) How many dogs have not learned any tricks?

Answer 1

0

Answer 2

The answer is 2.
From the set theory,
\[n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A.B)-n(B.C)-n(C.A) + n(A.B.C)\]
The dot implies intersection.
You need to add x to the R.H.S for dogs which did not learn any tricks.

Answer 3

or a venn worked from the inside out will be a good chart to fill in

Answer 4

Yeah venn diagrams are good. But they are very difficult to draw here.

Answer 5

Dogs in the GoodDog Obedience School win a blue ribbon for learning how to sit, a green ribbon for learning how to roll over, and a white ribbon for learning how to stay. There are 100 dogs in the school. 73 have blue ribbons, 39 have green ribbons, and 62 have white ribbons.

14 bgw

21-14 bg = 7
28-14 gw = 14
41-14 bw = 27

73 have blue ribbons
39 have green ribbons
62 have white ribbons.

Answer 6

adding up all the parts and subtracting the total number of participants leaves how many are not ribbon winners

Answer 7

Yeah,add all the parts and subtract from total number. correct.

Answer 8

if we try to reconstruct the original solution process:

(bgw) + (bgw-bg) + (bgw-bw) + (bgw-gw) + (b-bgw - bw - bg) + (g-bgw - gw - bg) + (w-bgw - bw - gw)

(bgw) + (bgw) -(bg+bw+gw) + (b+g+w) - (3bgw +2bw +2gw +2bg)

(2bgw) -(bg+bw+gw) + (b+g+w) - (3bgw +2bw +2gw +2bg)

(bgw) + (b+g+w) - 3(bw +gw +bg)

now to determine if ive messed up someplace lol

Answer 9

... personally, id stick with the venns for a set of 3 or less :)

Answer 10

the venn diagram probably is the simplest way :) label the unknown area as x and that is probably it

Answer 11

with a little thought to it: the parts remaining are:

b - bg - bw + bgw; since bgw is actually subtracted twice

14 bgw

21-14 bg = 7
28-14 gw = 14
41-14 bw = 27

73 - 21 - 41 + 14 = 25
39 - 21 - 28 + 14 = 4
62 - 28 - 41 + 14 = 7

100 - (14+7+14+27+25+4+7) = 100 - 98