Question 9 (0.5 points)

Service calls arriving at an electric company follow a Poisson distribution with an average arrival rate of 5656 per hour. Find

liberstina [14]

Answer:

The average number of service calls in a 15-minute period is of 14, with a standard deviation of 3.74.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval. The variance is the same as the mean.

Average rate of 56 calls per hour:

This means that $\mu = 56n$ , in which n is the number of hours.

Find the average and standard deviation of the number of service calls in a 15-minute period.

15 minute is one fourth of a hour, which means that $n = \frac{1}{4}$ . So

$\mu = 56n = \frac{56}{4} = 14$

The variance is also 14, which means that the standard deviation is $\sqrt{14} = 3.74$

The average number of service calls in a 15-minute period is of 14, with a standard deviation of 3.74.

4 0

2 years ago

Consider the polynomial p(x) = x^3 + 4x^2 + 6x − 36.

Nady [450]

Answer:

a. attached graph; zero real: 2

b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i

Step-by-step explanation:

p(x) = x³ + 4x² + 6x - 36

a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.

p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0

b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.

2 | 1 4 6 -36

1 6 18 0

x² + 6x + 18 = 0

Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²

√Δ = 6i

x = -6±6i/2 = 2(-3±3i)/2

x' = -3-3i

x" = -3+3i

p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i

4 0

3 years ago

A square is cut in half and the diagonal creating two equal triangles of each triangle has an area of 4.5 in.² what is the lengt

rusak2 [61]

8 Ft 9In i think dont kill me if its wrong

5 0

2 years ago

8. What is the result of (11x + 18i)(117 - 18i)?

Nonamiya [84]

1287x + 342 +(-198x + 2106) i

5 0

3 years ago

If the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week

serg [7]

The population after 20 weeks will be 403.42 $x_{0}$ in which $x_{0}$ is the initial population.

Given that the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week.

We are required to find the number of bacteria present after 10 weeks.

let the number of bacteria present at t is x.

So,

dx/dt∝x

dx/dt=kx

1/x dx=k dt

Now integrate both sides.

$\int\limits {1/x} \, dx$ = $\int\limits{k} \, dt$

log x=kt+log c----------1

Put t=0

log $x_{0}$ =0 +log c ( $x_{0}$ shows the population in beginning)

Cancelling log from both sides.

c= $x_{0}$

So put c= $x_{0}$ in 1

log x=kt+log $x_{0}$

log x=log $e^{kt}$ +log $x_{0}$

log x=log $e^{kt}x_{0}$

x= $e^{kt}x_{0}$

We have been given that the population triples in a week so we have to put the value of x=2 $x_{0}$ and t=1 to get the value of k.

2 $x_{0}$ = $e^{k} x_{0}$

2= $e^{k}$

log 2=k

We have to now put the value of t=20 and k=log 2 ,to get the population after 20 weeks.

x= $e^{20log 2}$ $x_{0}$

x= $e^{0.30*2}$ $x_{0}$

x= $e^{6}x_{0}$

x=403.42 $x_{0}$

Hence the population after 20 weeks will be 403.42 $x_{0}$ in which $x_{0}$ is the initial population.

Learn more about growth rate at brainly.com/question/25849702

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The given question is incomplete as the question incudes the following:

Calculate the population after 20 weeks.

5 0

2 years ago