Answer:
B) Independent; the 1st marble selection will not affect the 2nd marble selection.
Step-by-step explanation:
When finding the probability of events in mathematics, we have both independent and dependent events.
Independent events are events that occur when the results of selection of the first events does not affect the results or outcomes of the second events.
Dependent events are the opposite of Independent events. They are events that occur when the results or outcomes obtained from the second events is affected by the results or outcomes from the selection of the first events.
From the question, we can see that the first event is she picked one marble from the bag. The second event is she replaced the marble before picking another marble. By doing this, the total number of possible outcomes for the probabilities of both events remains the same and they are unaffected.
Therefore, we can say that the two events are Independent because the 1st marble selection will not affect the 2nd marble selection.
Answer:
Step-by-step explanation:
I really dont know but if I did I would say I'm a 6 grader
So to solve this question, your goal is to find out how the way it is solved is not correct.
Your answer would be: On the third line, the student adds the 8 to both sides instead of subtracting. The way the initial equation is given is
y-(-8)=-6(x-2). After distributing the six, the student should make the 8 positive because subtracting a negative makes a positive. After solving, the equation should look like: y(+8)=-6x+12, so you would subtract the 8 from both sides instead of adding it, and solve from there.
Answer:
The value is 
Step-by-step explanation:
From the question we are told that
The population proportion is 
The sample size is n = 563
Generally the population mean of the sampling distribution is mathematically represented as
Generally the standard deviation of the sampling distribution is mathematically evaluated as
=> 
=> 
Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as
Here 
is the sample proportion of persons with a college degree.
So
![P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )](https://tex.z-dn.net/?f=P%28%20-%20%280.05%20-%200.52%20%29%20%3C%20%20%5C%5E%20p%20%3C%20%20%280.05%20%2B%200.52%20%29%29%20%3D%20P%28%5Cfrac%7B%5B%5B0.05%20-0.52%5D%5D-%200.52%7D%7B0.02106%7D%20%3C%20%5Cfrac%7B%5B%5C%5Ep%20-%20p%5D%20-%20p%7D%7B%5Csigma%20%7D%20%20%3C%20%5Cfrac%7B%5B%5B0.05%20-0.52%5D%5D%20%2B%200.52%7D%7B0.02106%7D%20%29)
Here
![\frac{[\^p - p] - p}{\sigma } = Z (The\ standardized \ value \ of\ (\^ p - p))](https://tex.z-dn.net/?f=%5Cfrac%7B%5B%5C%5Ep%20-%20p%5D%20-%20p%7D%7B%5Csigma%20%7D%20%20%3D%20Z%20%28The%5C%20standardized%20%5C%20%20value%20%5C%20%20of%5C%20%20%28%5C%5E%20p%20-%20p%29%29)
=> ![P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]](https://tex.z-dn.net/?f=P%28%20-%20%280.05%20-%200.52%20%29%20%3C%20%20%5C%5E%20p%20%3C%20%20%280.05%20%2B%200.52%20%29%29%20%3D%20P%5B%5Cfrac%7B-0.47%20-%200.52%7D%7B0.02106%20%7D%20%20%3C%20%20Z%20%20%3C%20%5Cfrac%7B-0.47%20%2B%200.52%7D%7B0.02106%20%7D%5D)
=> ![P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]](https://tex.z-dn.net/?f=P%28%20-%20%280.05%20-%200.52%20%29%20%3C%20%20%5C%5E%20p%20%3C%20%20%280.05%20%2B%200.52%20%29%29%20%3D%20P%5B%20-2.37%20%3C%20%20Z%20%20%3C%202.37%20%5D)
=> 
From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is
and
So
=>
=>
=>
