According to a recent survey of adults, approximately 62% carry cash on a regular basis. The adults were also
1 answer:
According to the probabilities given, it is found that the correct option regarding the independence of the events is given by:
No, P(carry cash) != P(carry cash|have children).
<h3>What is the probability of independent events?</h3>If two events, A and B, are independent, we have that:
Which also means that:
In this problem, we have that:
- 62% carry cash on a regular basis, hence P(cash) = 0.62.
- 46% has children, hence P(children) = 0.46.
- Of the 46% who have children, 85% carry cash on a regular basis, hence P(cash|children) = 0.85.
Since P(carry cash) != P(carry cash|have children), they are not independent.
More can be learned about the probability of independent events at brainly.com/question/25715148
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Answer:
see below. The solution is the doubly-shaded area.
Step-by-step explanation:
Each boundary line will be dashed, because the "or equal to" case is <em>not included</em>. Each shaded area will be above the corresponding boundary line because the comparison symbol is y > .... That is, only y-values greater than (above) those in the boundary line are part of the solution.
Of course, the boundary lines are graphed in the usual way. Each crosses the y-axis at the value of the constant in its equation. Each has a slope (rise/run) that is the value of the x-coefficient in the equation.
