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3 years ago

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In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed

because the mobile phone may not receive the paging signal clearly. Consequently, the system will page a phone up to three times before giving up. If the results of all paging attempts are independent and a single paging attempt succeeds with probability 0.8, sketch a probability tree for this experiment and find the probability P[F] that the phone receives the paging signal clearly.

1 answer:

Alina [70]3 years ago

7 0

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒ $0.2\times 0.8$

⇒ $0.16$

On the third attempt the success probability will be:

⇒ $0.2\times 0.2\times 0.8$

⇒ $0.032$

So that the success probability will be:

⇒ $0.8 + 0.16 + 0.032$

⇒ $0.992$

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While discussing run-flat tires: Technician A says that some are self-sealing tires and are designed to quickly and permanently

musickatia [10]

Answer:

The correct option is d ( Neither A nor B)

Explanation:

Technician A made 2 mistakes in his statement.Firstly the tire is self supporting not self sealing.

Secondly, this tire does not provide permanent sealing of punctured area option a is incorrect.

This self-supporting tire after being affected with complete air leakage can temporarily bear the load of the car and avoid rolling over a distance of 80 km at a maximum speed of 55 mph. Here is what technician B suggested incorrectly as the tire after being.Here the technician B suggested incorrectly as the tire after being affected with puncture can not travel at any speed so option B is wrong

Since option a and b are incorrect and c is invalid.

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2 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

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The following is a list of metals and alloys:

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Answer:

A) Gray cast iron

B) Aluminum

C) Titanium alloy

D) Tool steel

E) Titanium alloy

F) magnesium

G) Tungsten

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2 years ago

What is the best way to submit your assignments?

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A. Email your teacher right away. It would be the safest option.

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3 years ago

Read 2 more answers

A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro

Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= $\frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt$

$= \int\limits^a_b {(5-3t)} \, dt$

$5t - \frac{3t^2}{2} +c$

v2 = 5x2 - 3x2 + c

= 10-6+c

= 4+c

$s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c$

S2 - S1

$=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)$

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

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2 years ago