Answer:
17
Step-by-step explanation:
You will have to subtract 36 and 19 and than you will get 17
Answer:
11 1/12
Step-by-step explanation:
6 1/3 = 19/3
7 1/4 = 29/4
2 1/2 = 5/2
19/3 + 29/4 - 5/2
We must find the LCM of 3, 4, and 2. This happens to be 12
3*4 = 12
4*3 = 12
2*6 = 12
Multiply each fraction by the factor that'll get it to 12.
19/3 * 4/4 = 76/12
29/4 * 3/3 = 87/12
5/2 * 6/6 = 30/12
Now go through the problem
76/12 + 87/12 - 30/12
76 + 87 = 163
163/12 - 30/12
163 - 30 = 133
133/12
Simplify
133/12 = 11.0833... or 11 1/12
Hope this helps.
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:
x=9
Step-by-step explanation:
It is 7*1+2=9
The output is 9
