At a fundraiser, a school group charges $8 for tickets for a "grab bag". You choose one bill at random from a bag that contains
1 answer:
Answer:
Probability of event = 0.28
Step-by-step explanation:
At a fundraiser, a school group charges $8 for tickets for a "grab bag".
You choose one bill at random from a bag that contains 36 $1 bills, 19 $5 bills, 6 $10 bills, 7 $20 bills, and 3 $100 bills.
The total number of bills = 36 + 19 + 6 + 7 +3 =71 bills.
We have to find the probability that i will be able to win enough to pay my ticket that is he must get at least $8.
The price would be sufficient if he gets $10, $20 or $100 bills.
So the favorable cases = 20
Total cases = 71
hence probability = = 0.28
The probability of him getting price greater than his ticket = 0.28
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Answer : option B
Parent Graph f(x) = is attached below
We analyze each option
A) f(x − 2) --> x - 2 shifts the graph 2 units to the right
B) f(x + 2) --> x + 2 shifts the graph 2 units to the left
C) f(x) − 2 --> f(x) - 2 shifts the graph 2 units down
D) f(x) + 2 --> f(x) + 2 shifts the graph 2 units up
In the graph of f(x)= , the x intercept at x=1
From the graph attached in the question, the x intercept at x=-1
Its means the parent graph is shifted 2 units to the left.
Option B: f(x + 2) --> x + 2 shifts the graph 2 units to the left
Hi there!
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I believe your answer is:
Option B)
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Here’s why:
- Most options (Second, Third, and Fourth) are in slope intercept form.
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- We are looking for the equation where '-5' is in b's place.
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Hope this helps you. I apologize if it’s incorrect.