The Figures shown are similar. What is the scale factor?

The number of requests for assistance received by a towing service is a Poisson process with rate θ = 4 per hour.a. Compute the

aliya0001 [1]

Answer:

a) 9.93% probability that exactly ten requests are received during a particular 2-hour period

b) 13.53% probability that they do not miss any calls for assistance

c) 2

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 time interval.

Poisson process with rate θ = 4 per hour.

This means that $\mu = 4n$ , in which n is the number of hours.

a. Compute the probability that exactly ten requests are received during a particular 2-hour period.

n = 2, so $\mu = 4*2 = 8$

This is P(X = 10). So

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

$P(X = 10) = \frac{e^{-8}*8^{10}}{(10)!} = 0.0993$

9.93% probability that exactly ten requests are received during a particular 2-hour period

b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?

n = 0.5, so $\mu = 4*0.5 = 2$

This is P(X = 0). So

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

13.53% probability that they do not miss any calls for assistance

c. How many calls would you expect during their break?

n = 0.5, so $\mu = 4*0.5 = 2$

4 0

3 years ago

Evaluate : limx→ tan2-sin2x x3

GrogVix [38]

Given:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Solve:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Use l'hopital's rule:

$\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}$

Simplify:

$\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}$

Apply the constant multiple rule:

$\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}$ $\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}$

Similary :

$\begin{gathered} =\frac{2\lim _{x\to0}(2\cos (2x)+12\tan ^4(2x)+16\tan ^2(2x)+4)}{3} \\ =\frac{2(6)}{3} \\ =4 \end{gathered}$

8 0

1 year ago

Vivi scored 5 goals in Saskatoon sticks lacrosse tournament. This was 1/8 of her teams goals. How many goals did vivi’s team sco

yKpoI14uk [10]

5/ (1/8) = 40. Therefore the team scored 40 goals

7 0

3 years ago

Which is true about rigid transformations and circles?

steposvetlana [31]

A rigid transformation is defined to be one in which the pre-image of the object and its new image after the transformation both have the exact same size and shape. So the answer in this question is:

<span>D. a circle's shape is preserved regardless of any rigid transformation</span>

5 0

3 years ago

Y=2+ square root of x-8 solve for x

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

y = 2√(x-8)

y² = 4(x-8) = 4x-32

4x = y²+32

x = 0.25y² +8

5 0

3 years ago