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alex41 [277]
3 years ago
10

Will give 12 points 2. At a fair, each person can spin two wheels of chance. The first wheel has the letters M and H. The second

wheel has the numbers 1, 2, and 3.
(a) List all the possible outcomes of the compound event.


(b) If you spin both wheels, what is the probability that you get either an M or a 2? Explain.
Mathematics
1 answer:
Nookie1986 [14]3 years ago
8 0
(a)  M1,  M2,  M3,  H1,  H2 and H3.

(b) possible outcomes are M1, M3, H2.

so probability = these  outcomes / all possible outcomes
= 3/6  = 1/2 Answer
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I need help on this question
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I believe this is pythag

15^2 x 11^2 = 27225

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165ft
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Niccola travels at an average speed of 40 mph for 50 miles.
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Answer:

36.67 [mile/hour].

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2 years ago
The table displays the number of students in each grade at Richmond High School who will or will not be attending the amusement
Pani-rosa [81]

Answer:

P( eleventh\: \: |\: \: FT)=  0.2453 \\\\P( eleventh\: \: |\: \: FT)=  24.53\% \\\\

Step-by-step explanation:

We are given a joint probability table.

There are four different graders in a school

1. Grade Ninth

2. Grade Tenth

3. Grade Eleventh

4. Grade Twelfth

Field trip refers to the students who will attending the amusement park field trip.

No field trip refers to the students who will not be attending the amusement park field trip.

We want to find out the probability that the selected student is an eleventh grader given that the student is going on a field trip.

P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)}

Where P(eleventh and FT) is the probability of students who are in eleventh grade and will be going to field trip

P(eleventh\:  and \: FT) = \frac{13}{92}  \\\\P(eleventh \:  and \: FT) = 0.1413

Where P(FT) is the probability of students who will be going to field trip

P(FT) = \frac{12}{92} + \frac{9}{92} + \frac{13}{92} + \frac{19}{92}\\\\P(FT) = 0.1304 + 0.0978 + 0.1413 + 0.2065\\\\P(FT) = 0.576 \\\\

So the required probability is

P( eleventh\: \: |\: \: FT)= \frac{P( eleventh\: \: and \: \: FT)}{P(FT)} \\\\P( eleventh\: \: |\: \: FT)= \frac{0.1413}{0.576} \\\\P( eleventh\: \: |\: \: FT)=  0.2453 \\\\P( eleventh\: \: |\: \: FT)=  24.53\% \\\\

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