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ioda
2 years ago
8

g Let Y denote the number of oil tankers arriving each day at a certain port and it is a Poisson random variable. Record shows t

hat the average of tankers arriving per day is 10. What is the probability that more than 7 tankers arrives in a certain day.
Engineering
1 answer:
Natalija [7]2 years ago
4 0

Answer:

0.77978

Explanation:

This is a Poisson distribution problem

Poisson distribution formula is given as

P(X = x) = (e^-λ)(λˣ)/x!

λ = mean = 10 tankers per day

x = variable whose probability is required

The probability that more than 7 tankers arrives in a certain day = 1 - (Probability that 7 or less tankers arrive in a certain day)

P(X > 7) = 1 - P(X ≤ 7)

P(X ≤ x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to x)

P(X ≤ 7) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + + P(X=6) + P(X=7) + P(X=8)

Computing this,

P(X≤7) = 0.22022

P(X > 7 ) = 1 - P(X≤7) = 1 - 0.22022 = 0.77978

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2.11 Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface
alexandr402 [8]

Answer:

The heat loss rate through one of the windows made of polycarbonate is 252W. If the window is made of aerogel, the heat loss rate is 16.8W. If the window is made of soda-lime glass, the heat loss rate is 1190.4W.

The cost associated with the heat loss through the windows for an 8-hour flight is:

For aerogel windows: $17.472 (most efficient)

For polycarbonate windows: $262.08

For soda-lime glass windows: $1,238.016 (least efficient)

Explanation:

To calculate the heat loss rate through the window, we can use a model of heat transmission by conduction throw flat wall. Using unidimensional Fourier law:

\frac{dQ}{dt}=\dot Q =-kS\nabla \vec{T}

In this case:

\dot Q =k\frac{S}{L} \Delta T

If we replace the data provided by the problem we get the heat loss rate through one of the windows of each material (we only have to change the thermal conductivities).

To obtain the thermal conductivity of the soda-lime glass we use the graphic attached to this answer (In this case for soda-lime glass k₃₀₀=0.992w/m·K).

To calculate the cost associated with the heat loss through the windows for an 8-hour flight we use this formula (using the heat loss rate calculated in each case):

Cost=C_{hc}\cdot \dot Q \cdot t \cdot n=1\frac{\$}{Kwh} \cdot \dot Q \cdot 8h \cdot 130

6 0
2 years ago
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea
Reika [66]

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

4 0
3 years ago
A poundal is the force required to accelerate a mass of 1 lbm at a rate of 1 ft/(s^2). Determine the acceleration of an object o
storchak [24]

Answer:

See it in the pic

Explanation:

See it in the pic

5 0
3 years ago
2. When manipulating your pedals, you should use your
astra-53 [7]

Answer:

D. left foot for the accelerator and your right foot for the brake.

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7 0
3 years ago
Read 2 more answers
A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos
Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0
2 years ago
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