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Scott plays a game using the spinner shown. During the game he will spin a total of 42 times. Which is the best prediction for the number of times he will get either "back 2" or "back 3" on the spinner? (all of the things you can land on are ({lose a turn}{forward 1}{back3}{forward2}{back1}{back2}
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how many spaces are labeled either "back 2" or "back 3"
1 space for all of the labels mentioned
so 2 spaces, out of how many total spaces?
6
the probability of landing on either "back 2" or "back 3" is 2/6 = 1/3
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multiply that with 42 and you will get your answer
So, \[\frac{ 1 }{ 3 }\ times \[\frac{42}{1}\
correct
42/3
reduce that
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14/1
so you should land on either space about 14 times
again notice how 14/42 = 1/3
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