Answer:
x >-8
Step-by-step explanation:
22−2x<38
Subtract 22 from each side
22-22−2x<38-22
-2x < 16
Divide each side by -2, remembering to flip the inequality
-2x/-2 > 16/-2
x >-8
Answer:
see explanation
Step-by-step explanation:
(a)
A recursive formula allows any term in the sequence to be found by adding the common difference d to the previous term.
Here d = - 4 , then recursive formula is
= 
- 4 with a₁ = 2
(b)
The explicit formula for an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 2 and d = - 4, thus
= 2 - 4(n - 1) = 2 - 4n + 4 = 6 - 4n ← explicit formula
(c)
Using the recursive formula
a₁ = 2
a₂ = 2 - 4 = - 2
a₃ = - 2 - 4 = - 6
Using the explicit formula
a₅ = 6 - 4(5) = 6 - 20 = - 14
a₁₀ = 6 - 4(10) = 6 - 40 = - 34
a₁₀₀ = 6 - 4(100) = 6 - 400 = - 394
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.
_____
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:
same
Step-by-step explanation:
