Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
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My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
8 is in the tenth position.
In order to round up, the number behind it (in this case. 9) must be one of the numbers of 5-9. Because 9 meets this requirement, you can round 8 to 9 and this will make 9 to 0.
Your answer is 490
Answer: EB is bisected by DF
A is the midpoint DF
EB is a segment bisector
FA=1/2FC.
Step-by-step explanation:
So to solve this you would make a tree with the first soup choices branching off into the other choices. Then when you wrote down all the possibilitys you would count all them and get your answer or you could just do 2*4*2=16
<span> so there would be 16 possibilitys </span>