Answer:
5
Step-by-step explanation:
The 32 that have blue and green ribbons include the 16 that have all three, so there are only 32 -16 = 16 that have only blue and green ribbons.
The 31 that have green and white ribbons likewise include the 16 with all three, so there are only 31 -16 = 15 that have only green and white ribbons.
The 38 that have blue and white ribbons include the 16 with all three, so there are only 38 -16 = 22 that have only green and white ribbons.
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If we add the numbers of blue, green, and white ribbons, we are counting twice the numbers that have 2 ribbons, and 3 times the numbers that have 3 ribbons. We want to count each kind of ribbon-holder only once. Hence the number of individual dogs with any number of ribbons is only ...
62 +55 +63 -(16 +15 +22) -2(16) = 95
Of the 100 dogs, 95 have ribbons, so 5 dogs have not learned any tricks.
W^2+5w-66=0
Factor
(w+11)(w-6)=0
w+11=0
w=-11 (not possible - cant have negative length
w-6=0
w=6
L(6)=66
L=11
Width = 6
Length = 11
The answer is 15%
hope it helps
please mark as brainliest
Answer:The answer has to be 9
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Step-by-step explanation: Because 2 plus 9 equals 11 and this is an addition problem
X= number of ounces in the full jar
3 = 1/6 x
multiply both sides by 6:
18=x
18 oz.
