Answer:
The Correct Answer On Edgen2020 Is
A) Linear pair postulate.
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Answer:
V = 3591.4 cm³
Step-by-step explanation:
Volume of Sphere = 
Where r = 9.5 cm
V = 
V = 
V = 
V = 3591.4 cm³
Answer:
yes
Step-by-step explanation:
yes it will
Answer:

And the z score for 0.4 is

And then the probability desired would be:

Step-by-step explanation:
The normal approximation for this case is satisfied since the value for p is near to 0.5 and the sample size is large enough, and we have:


For this case we can assume that the population proportion have the following distribution
Where:


And we want to find this probability:

And we can use the z score formula given by:

And the z score for 0.4 is

And then the probability desired would be:
