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Luba_88 [7]
3 years ago
14

A student was given an assignment to determine the percentage of students that brought a bag lunch to school. The student chose

5 random classrooms and surveyed every student in those classrooms. Which type of sampling did the student use?
A. convenience

B. systematic

C. stratified

D. cluster
Mathematics
2 answers:
tiny-mole [99]3 years ago
8 0
I believe it's cluster sampling.
JulsSmile [24]3 years ago
3 0

Answer: D. cluster

Step-by-step explanation:

A cluster sampling is a random sampling method in which a researcher splits the entire population into separate groups known as clusters.Then, he draw a random sample of clusters from the population.

After this he performs his analysis on data from the sampled clusters.

Given: A student was given an assignment to determine the percentage of students that brought a bag lunch to school. The student chose 5 random classrooms and surveyed every student in those classrooms.

Here classes is denoted as clusters and the student as researcher.

Hence, the cluster sampling is used by the student.

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3 years ago
A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can cost
Advocard [28]

Answer:

Radius = 1.12 inches and Height = 4.06 inches

Step-by-step explanation:

A soup can is in the shape of a right circular cylinder.

Let the radius of the can is 'r' and height of the can is 'h'.

It has been given that the can is made up of two materials.

Material used for side of the can costs $0.015 and material used for the lids costs $0.027.

Surface area of the can is represented by

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Now the function that represents the cost to construct the can will be

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Now we place the value of h in the equation (1) from equation (2)

C=0.054\pi r^{2}+0.03\pi r(\frac{16}{\pi r^{2}})

C=0.054\pi r^{2}+0.03(\frac{16}{r})

C=0.054\pi r^{2}+(\frac{0.48}{r})

Now we will take the derivative of the cost C with respect to r to get the value of r to get the value to construct the can.

C'=0.108\pi r-(\frac{0.48}{r^{2} })

Now for C' = 0

0.108\pi r-(\frac{0.48}{r^{2} })=0

0.108\pi r=(\frac{0.48}{r^{2} })

r^{3}=\frac{0.48}{0.108\pi }

r³ = 1.415

r = 1.12 inch

and h = \frac{16}{\pi (1.12)^{2}}

h = 4.06 inches

Let's check the whether the cost is minimum or maximum.

We take the second derivative of the function.

C"=0.108+\frac{0.48}{r^{3}} which is positive which represents that for r = 1.12 inch cost to construct the can will be minimum.

Therefore, to minimize the cost of the can dimensions of the can should be

Radius = 1.12 inches and Height = 4.06 inches

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