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3 years ago

What is the factored form of 3x + 21x-24?

Mathematics

1 answer:

anastassius [24]3 years ago

4 0

Answer:

factorize form of the equation is $24(x-1)=0$ , where $x=1$

Step-by-step explanation:

Given Equation:

$3x + 21x-24=0$

Adding the terms of 'x' and taking constants on the other side.

$24x-24=0$

The sign of the constant will change on changing the side of it.

$24x=24$

$x=\frac{24}{24}$

$x=1$

So, the value of 'x' is 1

The factorized form of the equation can be made by taking the common value from the equation i.e

$24x-24=0\\$

$24(x-1)=0$

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4 years ago

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

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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.

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

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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:

$P(X > x) = 0.08$