Answer:
12). LM = 37.1 units
13). c = 4.6 mi
Step-by-step explanation:
12). LM² = 23² + 20² - 2(23)(20)cos(119)°
LM² = 529 + 400 - 920cos(119)°
LM² = 929 - 920cos(119)°
LM =
=
= 37.08
≈ 37.1 units
13). c² = 5.4² + 3.6² - 2(5.4)(3.6)cos(58)°
c² = 29.16 + 12.96 - 38.88cos(58)°
c² = 42.12 - 38.88cos(58)°
c =
c =
c = 4.6386
c ≈ 4.6 mi
Answer:
The margin of error is of 0.73 oz.
The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
That is z with a pvalue of , so Z = 2.575.
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The margin of error is of 0.73 oz.
The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.
The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.
The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.