At a fundraiser, a school group charges $6 for tickets for a "grab bag." You choose one bill at random from a bag that contains
39 $1 bills, 19 $5 bills, 3 $10 bills, 7 $20 bills, and 3 $100 bills. Is it likely that you will win enough to pay for your ticket?
2 answers:
Answer:
124
Step-by-step explanation:
Answer:
no, the probability is only about 18%
Step-by-step explanation:
there are total of 71 bills of varying denominations in the bag
you need to select a bill that is $10 or more to cover your $6 expense; therefore, P($ ≥ 10) = 3 + 7 + 3 / 71 or 13 /71, about 18%
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Answer:
the height of the front doorstep is 12 cm
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102 because it’s kind of like a 180 but not I’m direct form
Answer:
Odd integers are always one more that an even integer:
1st: 2n+1
2nd: 2n+3
Step-by-step explanation:
14n+7 = 10n+15
4n = 8
n = 2
------------
1st: 2n+1 = 5
2nd: 2n+3 = 7
7x=5*(x+2)
7x=5x+10
7x-5x=10
2x=10
x=5
Answer:
Your answer should be B.
