nikiwold963526 nikiwold963526

03-05-2021
Mathematics

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At a fundraiser, a school group charges $7 for tickets for a "grab bag." You choose one bill at random from a bag that contains 43 $1 bills, 22 $5 bills, 7 $10 bills, 6 $20 bills, and 1 $100 bill. Is it likely that you will win enough to pay for your ticket?

It is likely that you will win enough to pay for your ticket because the probability of winning enough to pay for your ticket as a simplified fraction is what ___

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charlesstark2000 charlesstark2000

03-05-2021

Answer:

Step-by-step explanation:

All that greater than 7$ can pay for the ticket

thus

%=(7+6+1)/(43+22+7+6+1)=17.72151899%

(keep sig-figs as you need)

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