A confidence interval tells us how many percents we are confident about the range of a parameter. In this problem, <span>a 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). That means we're 95% confident that the Americans spend from 1.38 hours to 1.92 hours per day on average relaxing or pursuing activities they enjoy. In other words, 95% of the samples of the same size would have a mean number of hours relaxing or pursuing activities they enjoy between 1.38 to 1.92.</span>
<u>Answer</u>
2,268
<u>Explanation</u>
<u>
</u>
<u>By grouping 378 ca</u>n be written as,
378 = 300 + 70 + 8
6× 378 = 6 × (300 + 70 + 8)
= (6×300)
+ (6×70) + (6×8)
= 1800
+ 420 + 48
= 2,268
9514 1404 393
Answer:
132°
Step-by-step explanation:
Name the vertex of each angle the same as the angle letter. Name the intersection of the "horizontal" and "vertical lines" point Q.
Angle ZXQ is vertical to ∠x, so is the same measure.
Angle YQX is the value that makes the sum of angles in triangle XYQ be 180°. That is ...
∠YQX = 180° -51° -57° = 72°
This is also the measure of its vertical angle in the other triangle. Angle z is the sum of that vertical angle and 60°, so we have ...
∠z = 72° +60°
∠z = 132°
_____
<em>Additional comment</em>
The relations we used are ...
- vertical angles are congruent
- sum of angles in a triangle is 180°
- an exterior angle is equal to the sum of the remote interior angles
The nearest tenth is still zero. Reason being you have to start from the right, and once you round the far most 9 to the zero it turns to a one which is too small of a number to round up.
Answer:
a) 0.1558
b) 0.7983
c) 0.1478
Step-by-step explanation:
If we suppose that small aircraft arrive at the airport according to a <em>Poisson process</em> <em>at the rate of 5.5 per hour</em> and if X is the random variable that measures the number of arrivals in one hour, then the probability of k arrivals in one hour is given by:
(a) What is the probability that exactly 4 small aircraft arrive during a 1-hour period?
(b) What is the probability that at least 4 arrive during a 1-hour period?
(c) If we define a working day as 12 hours, what is the probability that at least 75 small aircraft arrive during a working day?
If we redefine the time interval as 12 hours instead of one hour, then the rate changes from 5.5 per hour to 12*5.5 = 66 per working day, and the pdf is now
and we want <em>P(X ≥ 75) = 1-P(X<75)</em>. But
hence
P(X ≥ 75) = 1-0.852 = 0.1478
