Point S is on line segment R T ‾ RT . Given R S = 4 x − 10 , RS=4x−10, S T = 2 x − 10 , ST=2x−10, and R T = 4 x − 4 , RT=4x−4, d
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Answer:
<em>Probability that the spinner will stop on an odd number or a multiple of 5 is </em><em>0.6</em>
Step-by-step explanation:
Probability =
We are given the equal regions numbered from 1 through 20 which means that our total possible outcomes are 20
<em>Total possible outcomes: 20</em>
<em>Outcomes that spinner will stop on an odd number, n(Odd): 10</em>
1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Probability of spinner stoping on Odd number:
P(Odd) = =
=
= 0.5
Outcomes that spinner will stop on a multiple of 5, n(5): 4
5, 10, 15, 20
Probability of spinner stoping on multiple of 5:
P(5) = =
=
= 0.2
Odd numbers which are a multiple of 5 are: 5 and 15
So,
P(Odd and 5) =
Thus Probability of spinner stopping at odd number or a multiple of 5 becomes:
P(Odd or 5) = P(Odd) + P(5) - P(Odd and 5) = 0.5 + 0.2 - 0.1 = 0.6