Answer:
(C) 7/2
Step-by-step explanation:
slope = y2-y1/x2-x1
Answer:
The supplier becomes less accurate than they otherwise would have tried to claim. A further explanation is below.
Step-by-step explanation:
According to the provider, this same width of that same confidence interval would be as follows:
= 
= 
Depending on the input observed, the width including its confidence interval would be as follows:
= 
= 
As even the width of that interval again for survey asserted > the width including its confidence interval according to the provider's statement, we could conclude that such is the appropriate reaction.
Calculate the z-score for the given data points in the item using the equation,
z-score = (x - μ) / σ
where x is the data point, μ is the mean, and σ is the standard deviation.
Substituting,
(47.7) z-score = (47.7 - 52.5)/2.4 = -2
This translates to a percentile of 2.28%.
(54.9) z-score = (54.9 - 52.5)/2.4 = 1
This translates to a percentile of 84.13%.
Then, subtract the calculate percentiles to give us the final answer of <em>81.85%.</em>
Thus, 81.85% of the Siberian Husky sled dogs are expected to weigh between 47.7 and 54.9 lbs.
1/6 for the first question and 1/3 for the second
For the first 6 1/3
second is 14/25
third 2 29/42