Answer:
Step-by-step explanation:
The absolute change in game length is ...
(180 min) -(150 min) = 30 min . . . . change in game length
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The relative change is expressed as a fraction of the original game length:
(30 min)/(150 min) × 100% = 20% . . . . relative change in game length
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.)
Answer:
48
Step-by-step explanation:
Since there is an absolute value, the answer must be positive. Multiply 2 by 8 since two negatives equal a positive. So, |15-47 + 9 - 8 -17|. Then add and subtract to get the answer. -32 + 9 - 8 -17 = -48. |-48| = 48
The first step we take is to factor out a GCF:
2d(d³ + 3d² - 9d - 27)
Now, we can factor what's in the parenthesis by grouping (don't forget to keep the 2d we factored out):
2d[d²(d+3) -9(d+3)]
2d(d²-9)(d+3)
d²-9 can still be factored because it is the difference of two squares:
2d(d+9)(d-9)(d+3)
That is the completely factored form.
Answer:
123
Step-by-step explanation:
