The discharge of suspended solids from a phosphate mine is normally distributed with mean daily discharge 27 milligrams per liter (mg/l) and standard deviation 14 mg/l. in what proportion of the days will the daily discharge be less than 13 mg/l?

To solve this problem we use the z test. The formula for z score is:

z = (x – u) / s

where,

x = sample value = 13 and below

u = mean value = 27

s = standard deviation = 14

 

z = (13 – 27) / 14 = -1

 

using the z table to get the standard normal probabilities, the p value at z = -1 is:

P = 0.1587

 

This means that 15.87% that the discharge is 13 mg/L and below.

The solution to the problem is as follows:

The force of friction can be calculated using the equation F = uN, where F is the magnitude of the force, u is the coefficient of friction, and N is the normal force. For an inclined plane at 18 degrees above the horizontal, the normal force is given by N = mg cos theta. Putting all the equations together,

F = umg cos theta
F = (0.09)(1250)(9.80) cos(18)
F = 1.05 kN

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