Based on data collected from metal shredders across the nation, the amount L of extractable lead in metal shredder residue has an approximate exponential distribution with mean μ = 2.0 milligrams per liter. What is the probability that L is greater than 1.20 milligrams per liter?

The cumulative distirbution function is given by:

And using the complement rule we have:

Step-by-step explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

Solution to the problem

For this case we define the random variable L="the amount L of extractable lead in metal shredder residue" and the distirbution for X is given by:

For this case we want this probability:

The cumulative distirbution function is given by:

And using the complement rule we have:

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

24

step-by-step explanation:

if it takes 8 pounds to make 12 divide 12 by 8 and you get 1.5, 1.5 x 16 = 24