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3 years ago

Which ordered pair is in the solution set of the system of linear inequalities

Mathematics

2 answers:

kompoz [17]3 years ago

6 0

No solution
because a number can not be less than and greater than the same number.

Monica [59]3 years ago

5 0

Answer:

Option D, no solution

Step-by-step explanation:

The system of linear inequalities is

y > $\frac{3}{2}$ (x) - 1 ------(1)

y < $\frac{3}{2}$ (x) - 1 ------(2)

Inequality (1) has all the ordered pairs in the solution set above the line

y = $\frac{3}{2}$ (x) - 1

and inequality (2) has all the ordered pairs in the solution set below the line

y = $\frac{3}{2}$ (x) - 1

Therefore, solution set of these two inequalities will no be common.

Option D, no solution is the answer.

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2 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

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a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

$\ln{e^{-0.01x}} = \ln{0.08}$

$-0.01x = \ln{0.08}$

$x = -\frac{\ln{0.08}}{0.01}$

$x = 252.6$

Capacity of 252.6 cubic feet per second

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3 years ago