<u><em>Answer:</em></u>
0.3
<u><em>Explanation:</em></u>
We are given that:
Probability that a randomly chosen person lives in the East is 0.12
P(East) = 0.12 ..................> I
Probability that a randomly chosen person lives in the West is 0.18
P(West) = 0.18 ............> II
Now, we want to find the probability that the randomly chosen person lives in the East or West
<u>To do this, we will use the following formula:</u>
P(A or B) = P(A) + P(B)
<u>Applying this rule to our givens:</u>
P(East or West) = P(East) + P(West)
P(East or West) = 0.12 + 0.18
P(East or West) = 0.3
Hope this helps :)
Answer:
23 and 33
Step-by-step explanation:
23+33=56
33-23=10
Answer:
30 + 6
Step-by-step explanation:
the product of 9 & 4 is 36,
so, 30 + 6 is 36.
) It bends light rays inward.
Using the z-distribution, it is found that the 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).
<h3>z-distribution interval:</h3>
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
- In which z is the z-score that has a p-value of
.
For this problem:
- 1215 samples, hence
.
- 33% was mislabeled or misidentified, hence
.
- 95% confidence level, hence
, z is the value of Z that has a p-value of 
, so 
.
<h3>The lower limit of this interval is:</h3>
<h3>The upper limit of this interval is:</h3>
The 95% confidence interval for the proportion of all seafood sold in the United States that is mislabeled or misidentified is (0.3036, 0.3564).
You can learn more about the use of the z-distribution to build a confidence interval at brainly.com/question/25730047
