A. The mean and standard deviation.
The mean of a sampling distribution is approximately equal to the mean of the population. Given that the mean of the population is equal to 174.5, the mean of the sampling distribution is also this value.
The standard deviation of a sample distribution is equal to,
u(m) = u/sqrt n
Substituting the known values,
u(m) = 6.9 / sqrt 25 = 1.38
b. Get the z-score of both items,
z-score = (data point - mean) / standard deviation
z-score of 172.5
z-score = (172.5 - 174.5) / 1.38 = -1.49
This translates to 0.068.
z-score of 175.8
z-score = (175.8 - 174.5) / 1.38 = 0.94
This translates to 0.83.
The difference between the two z-scores is 0.762.
The number of samples with this height is 0.762(200) which is equal to approximately 152.
c. z-score of 172 centimeters
z-score = (172 - 174.5) / 1.38
z-score = -1.81
This translates to 0.03.
The number of people with this height from the sample is (0.03)(200) = 6
Answer: A or B
Step-by-step explanation:
The volume of a cylinder can be found using the formula:
π r² h,
where r is the radius of the circular base and h is the height of the cylinder.
If we plug in the measurements of the cylinder, we get:
π (6²) (28)
When this is simplified, we get that the volume of the cylinder is:
1008π cubic cm
Thus, if each marble has a volume of 36π cubic cm, then to find how many marbles will fit into the vase we must divide the vases total volume by the volume of each marble.
1008π / 36π = 28
Therefore, the answer is D. 28 marbles
To find the area of his exclusion zone you would need to understand that a triangle with dimensions of 3, 4, and 5 represent a right triangle.
This means the exclusion zone would be applied to the base and the height of the triangular space.
You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.
Please see the attached picture to understand this.
You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.
(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².
The exclusion space is 9 km².