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
Step-by-step explanation:
slope = y2 - y1
__________
x2 - x1
slope = (-10 -2) / (-16 - 4)
slope = -12/-20
slope = 3/5
Answer:
Graph A = 3
Graph B = 0
Graph C = 1
Graph D = 2
Step-by-step explanation:
Solutions are where the lines intersect
295×34=10030 toys in 34 days
Answer:
20x^2−2x−6
Step-by-step explanation:
(4x+2)(5x+−3)
=(4x)(5x)+(4x)(−3)+(2)(5x)+(2)(−3)
=20x2−12x+10x−6
=20x2−2x−6