9514 1404 393
Answer:
6 : 5
Step-by-step explanation:
The ratio is ...
dogs : cats = 42 : 35 = (7·6) : (7·5) = 6 : 5
The ratio of dogs to cats at the shelter is 6 to 5.
The answer is a.
4 P 4 = 24
Answer:
![f(g(x))=\frac{1}{(x^{2}+1)^{2}} +\sqrt[3]{x^{2}+1}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D%5Cfrac%7B1%7D%7B%28x%5E%7B2%7D%2B1%29%5E%7B2%7D%7D%20%2B%5Csqrt%5B3%5D%7Bx%5E%7B2%7D%2B1%7D)
Step-by-step explanation:
we have
![f(x)=x^{2} +\frac{1}{\sqrt[3]{x}}](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E%7B2%7D%20%2B%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D)
we know that
In the function
The variable of the function f is now the function g(x)
substitute
![f(g(x))=(\frac{1}{x^{2}+1})^{2} +\frac{1}{\sqrt[3]{(\frac{1}{x^{2}+1})}}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D%28%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%2B1%7D%29%5E%7B2%7D%20%2B%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7B%28%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%2B1%7D%29%7D%7D)
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:
I think the picture here explains you everything
hope this helped
