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Dave's sister baked $3$ dozen pies of which half contained chocolate, two thirds contained marshmallows, three-fourths contained cayenne, and one-sixths contained salted soy nuts. What is the largest possible number of pies that had none of these ingredients?

Respuesta :

merlynthewhizz
Number of pies = 3×12 = 36

Pies contained chocolate = [tex] \frac{1}{2} [/tex]×36 = 18
Pies contained marshmallows = [tex] \frac{2}{3} [/tex]×36 = 24
Pies contained cayenne = [tex] \frac{3}{4} [/tex]×36 = 27
Pies contained soy nut = [tex] \frac{1}{6} [/tex]×36 = 6

Pies that don't contain any of these 36-(18+24+27+6) = 25 pies

Answer:

its 9

Step-by-step explanation:

im using aops answer said 9

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