Which condition would prove ΔDEF ~ ΔJKL? A. EF:KL = 1.5:1
B. DE:JK = 1:1.5
C. DF:JL = 1.5:1
D. DF:JL = 1:1.5

go and ask your mom to you

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: