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yaroslaw [1]
3 years ago
11

A cylindrical can is to have volume 1500 cubic centimeters. determine the radius and the height which will minimize the amount o

f material to be used. note that the surface area of a closed cylinder is s=2rh+2r2 and the volume of a cylindrical can is v=r2h.
Mathematics
1 answer:
Degger [83]3 years ago
8 0
Assuming R and H:
So volume is pir^2 * H = 1500 and H = 1500/(pir^2)  while surface area is A= 2pir*H + 2pir^2 
A = 2pir(r+h)= 2piR^2 + 2pir*1500/(pir^2)= 2piR^2 + 3000/r
For A to take minimum, get the derivative 4pir - 3000/R^2 and let it be 0 
4pir^3 - 3000 = 0 
r = cbrt(3000/(4pi)) ≈ 6.20
h = 1500/(pi(6.20)^2) ≈ 12.42
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We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of  length of human pregnancies is a bell shaped distribution that is a normal distribution.

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b) a) P(last between 236 and 296)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least 1-\dfrac{1}{k^2}  data lies within k standard deviation of mean.

For k = 2

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Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

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