Given that Bill
is on the school archery team. the target has a center bull's-eye and
two rings around the bull's-eye.
Given the table below that gives the probabilities of outcomes.

The probability <span>that Bill will get the next arrow in the inner or outer ring is given by
P(inner ring) + P(outer ring) = 0.297 + 0.423 = 0.72</span>
Therefore, <span>the probability that Bill will get the next arrow in the inner or outer ring is 0.72</span>
7(h-4) = 2h + 17 distribute to get:
7h - 28 = 2h + 17 subtract 2h from both sides to get:
5h - 28 = 17 add 28 to both sides to get:
5h = 45 divide 5 from both sides to get:
h = 9
The probability according to the given scenario is:
(a) <em>0.0395</em>
(b) <em>0.95</em>
(c) <em>0.076</em>
The given values are:
(a)
The probability that the b/w has tuberculosis as well as having (+) test will be:
→ 
By substituting the values, we get
→ 
→ 
(b)
The probability that a person does not have tuberculosis will be:
→ 
→ 
→ 
(c)
The person does not have tuberculosis as well as have a (+) test, the probability will be:
→ 
→ 
→ 
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Answer:
is this all?
Step-by-step explanation: