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denpristay [2]
2 years ago
15

The time it takes a printer to print a job is an Exponential random variable with the expectation of 12 seconds. You send a job

to the printer at 10:00 am, and it appears to be third in line. What is the probability that your job will be ready before 10:01
Mathematics
1 answer:
Mnenie [13.5K]2 years ago
8 0

Answer:

0.8753

Step-by-step explanation:

Calculate the probability that your job will be ready before 10.01 am

Here, the parameter of an Exponential is E(X)=12

Now, to calculate the third job probability, it follows Poisson Distribution with parameter 1/λ

Therefore, E(Y) =1/12

Here, The third job will be ready for 10:01 AM, then E(Y)=61/12

Therefore, the required probability is

P(X\geq 3)=1-P(X

=1- POISSON(3,5,true)

=1-0.1246

=0.8753

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Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to
nikklg [1K]

Answer:

Part 1)

See Below.

Part 2)

\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)

Step-by-step explanation:

Part 1)

The linear approximation <em>L</em> for a function <em>f</em> at the point <em>x</em> = <em>a</em> is given by:

\displaystyle L \approx f'(a)(x-a) + f(a)

We want to verify that the expression:

1-36x

Is the linear approximation for the function:

\displaystyle f(x) = \frac{1}{(1+9x)^4}

At <em>x</em> = 0.

So, find f'(x). We can use the chain rule:

\displaystyle f'(x) = -4(1+9x)^{-4-1}\cdot (9)

Simplify. Hence:

\displaystyle f'(x) = -\frac{36}{(1+9x)^{5}}

Then the slope of the linear approximation at <em>x</em> = 0 will be:

\displaystyle f'(1) = -\frac{36}{(1+9(0))^5} = -36

And the value of the function at <em>x</em> = 0 is:

\displaystyle f(0) = \frac{1}{(1+9(0))^4} = 1

Thus, the linear approximation will be:

\displaystyle L = (-36)(x-(0)) + 1 = 1 - 36x

Hence verified.

Part B)

We want to determine the values of <em>x</em> for which the linear approximation <em>L</em> is accurate to within 0.1.

In other words:

\displaystyle \left| f(x) - L(x) \right | \leq 0.1

By definition:

\displaystyle -0.1\leq f(x) - L(x) \leq 0.1

Therefore:

\displaystyle -0.1 \leq \left(\frac{1}{(1+9x)^4} \right) - (1-36x) \leq 0.1

We can solve this by using a graphing calculator. Please refer to the graph shown below.

We can see that the inequality is true (i.e. the graph is between <em>y</em> = 0.1 and <em>y</em> = -0.1) for <em>x</em> values between -0.179 and -0.178 as well as -0.010 and 0.012.

In interval notation:

\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)

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