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2 years ago

Please help thank you

Mathematics

1 answer:

ioda2 years ago

6 0

Tan65°=h/75m

2.1445 × 75 = h

h = 160.83

Step-by-step explanation:

This is very very simple

You might be interested in

Assume that power surges occur as a Poisson process with rate 3 per hour. These events cause damage to a computer, thus a specia

lidiya [134]

Answer:

a) 0.30327

b) 0.07582

Step-by-step explanation:

This is a Poisson distribution problem

Poisson distribution probability function is given as

P(X = x) = (e^-λ)(λˣ)/x!

where λ = mean = 3 power surges per hour

x = variable whose probability is required

a) If a single disturbance in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of a single disturbance (power surge) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 1 disturbance per 10 minutes

P(X = 1) = (e^-0.5)(0.5¹)/1! = (e⁻⁰•⁵)(0.5)/1!

= 0.30327

b) If two disturbances in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of two disturbances (power surges) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 2 disturbance per 10 minutes

P(X = 2) = (e^-0.5)(0.5²)/2! = (e⁻⁰•⁵)(0.25)/2!

= 0.07582

Hope this Helps!!!

3 0

3 years ago

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.