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Salsk061 [2.6K]

3 years ago

6

Consider an espresso stand with a single barista. Customers arrive at the stand at the rate of 28 per hour according to a Poisso

n distribution. Service times are exponentially distributed with a service rate of 35 customers per hour. The probability that the server is busy is:

1 answer:

adoni [48]3 years ago

4 0

Answer:

The probability that the server is busy is P=0.56.

Step-by-step explanation:

We have a quieing theory problem with M/M/1

In queing theory, the probability of the server being busy can be expressed as:

$P=\frac{\lambda}{\lambda+\mu}$

being μ: the time between services and λ: the time between customers arrival.

Then we can calculate:

$P=\frac{\lambda}{\lambda+\mu}=\frac{\frac{1}{28} }{\frac{1}{28}+\frac{1}{35} } =\frac{0.036}{0.036+0.029}=0.56$

The probability that the server is busy is P=0.56.

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Solve without using a calculator:<br><br> sin²20° + sec²20°

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1

as

1

2

.

sin

2

(

20

°

)

+

1

2

cos

2

(

20

°

)

Rewrite

1

2

cos

2

(

20

°

)

as

(

1

cos

(

20

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2

.

sin

2

(

20

°

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+

(

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2

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20

°

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3 years ago

In the town zoo 3/28 of the animals are birds of the birds for 4/15 are birds of prey what fraction of the animals at the zoo ar

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7 0

4 years ago

Read 2 more answers

(10 pts) A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstruc

dimulka [17.4K]

Answer:

309

Step-by-step explanation:

To determine the estimated free-flow speed

$FFS = 75.4 -f_{LW}-f_{LC} - 3.22 TRD^{0.84}$

From the table "Adjustment for lane width," which corresponds to a lane width of 12 feet. As a result, $f_{LW}$ equals 0 mi/h.

Take the meaning from the table "Adjustment for right-shoulder lateral clearance," which corresponds to 6 feet of right shoulder lateral clearance and two lanes in one direction.

The $f_{LC} = 0 mi/h$

$FFS = 75.4 -0-0-3.22 ( \dfrac{5}{6})^{0.84}$

$= 72.64 \ mi/h$

$\simeq 73 \ mi/h$

Determine the peak-hour factor

$PHF = \dfrac{V}{V_{15}\times 4} \\ \\ =\dfrac{1800}{700\times 4}\\ \\ = 0.6429$

Now, Find the heavy-vehicle adjustment factor.

$v_P = \dfrac{V}{PHF\times N \times f_{HV}\times f_P} --- (1)$

Take the value for the 15-minute passenger car equivalent flow rate from the table "LOS requirements for freeway segments" for the FFS and LOS C conditions. The free-flow speed is estimated to be 73 miles per hour

$v_p = 1735+ \dfrac{73-70}{75-70}(1775-1735)$

$v_p = 1759 \ pc/h/In$

Think about for familiar users the value of f_p = 1.00

Replace all of the values obtained in (1)

$v_p = \dfrac{V}{PHF \times N\times f_{HV}\times f_p}$

$1759 = \dfrac{1800}{0.6429\times 2 \times f_{HV}\times 1}$

$f_{HV} = \dfrac{1800}{0.6429\times 2 }\times \dfrac{1}{1759}$

= 0.795

Calculate the percentage of trucks and buses in the flow of traffic stream.

$f_{HV} = \dfrac{1}{1+P_{\tau}(E_{\tau}-1) + P_R(E_R-1)}$

Take the values from the table "Passenger car equivalent for extended highway segments" referring to trucks and buses and rolling terrains for passenger car equivalent for trucks and buses and recreational vehicles. As a result, $E_T$ has a value of 2.5 and $E_R$ has a value of 2.0.

Since $P_R$ is zero and there is no recreational vehicle.

Then;

$0.759 = \dfrac{1}{1+ P_T(2.5 -1) +0}$

$P_T = 0.1719$

Finally, to estimate the maximum number of large trucks and buses; we have:

$= V\times P_T = 1800 \times 0.1719$

Maximum number of large trucks and buses = 309

4 0

3 years ago