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tangare [24]
2 years ago
6

A worker at a landscape design center uses a machine to fill bags with potting soil. Assume that the quantity put in each bag fo

llows the continuous uniform distribution with low and high filling weights of 14.1 pounds and 16.2 pounds, respectively.Calculate the expected value and the standard deviation of this distribution.
Mathematics
1 answer:
ddd [48]2 years ago
4 0

Answer:

Mean = 15.15

Standard deviation = 0.6062

Step-by-step explanation:

We are given a uniform distribution of  quantity to be put in each bag.

Low and high filling weights:

14.1 pounds and 16.2 pounds

a = 14.1

b = 16.2

a) Mean:

\mu = \displaystyle\frac{a+b}{2}\\\\\mu = \frac{14.1+16.2}{2} = 15.15

b) Standard Deviation:

\sigma = \sqrt{\displaystyle\frac{(b-a)^2}{12}}\\\\= \sqrt{\dfrac{(16.2-14.1)^2}{12}} = \sqrt{0.3675} = 0.6062

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For what value of k does the equation 6(x + 1) + 2 = 3(k5x + 1) + 3 have no solution?
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Answer:

k = (6/15)

Step-by-step explanation:

The equation is:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

To have no solutions, we need to have something like:

x + 7 = x + 4

where we can remove x in both sides and end with

7 = 4

So this equation is false, meaning that there is no value of x such that this equation is true, then the equation has no solutions.

First, let's try to simplify our equation:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

6*x + 6 + 2 = 3*k*5*x + 3*1 + 3

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