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Serga [27]
3 years ago
10

The lengths of pregnancies in a small rural village are normally distributed with a mean of 264.1 days and a standard deviation

of 12.9 days. In what range would you expect to find the middle 95% of most pregnancies
Mathematics
1 answer:
KiRa [710]3 years ago
5 0

Answer:

The range of the 95% data (X) =  238.3 days <  X < 289.9 days

Step-by-step explanation:

Given;

mean of the normal distribution, m = 264.1 days

standard deviation, d = 12.9 days

between two standard deviation below and above the mean is 96% of all the data.

two standard deviation below the mean = m - 2d

                                                                   = 264.1 - 2(12.9)

                                                                   = 238.3 days

two standard deviation above the mean = m + 2d

                                                                    = 264.1 + 2(12.9)

                                                                     = 289.9 days

The middle of the 95% of most pregnancies would be found in the following range;

238.3 days <  X < 289.9 days

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Subtract 1/2n from both sides:

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Step-by-step explanation:

This question is incomplete; here is the complete question.

A closed cylindrical can of fixed volume V has radius r. (a) Find the surface area, S, as a function of r. (b) What happens to the value of S approaches to infinity? (c) Sketch a graph of S against r, if  V=10 cm³.

A closed cylindrical can of volume V is having radius r and height h.

a). Surface area of a cylinder is given by

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S = 2πr² + 2πrh

S = 2πr(r + h)

b). Since surface area is directly proportional to radius of the can

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From the formula V = πr²h

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By placing the value of h in the formula of surface area,

S = 2\pi r(r+\frac{10}{\pi r^{2}})

Now we can get the points to plot the graph,

r       -2             -1         0       1            2

S    -13.72     -13.72     0    26.28    35.13

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