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3 years ago

13

Network breakdowns are unexpected rare events that occur every 3 weeks, on the average. Compute the probability of more than 4 b

reakdowns during a 21-week period

1 answer:

allsm [11]3 years ago

5 0

Answer:

0.827

Step-by-step explanation:

Data provided in the question:

Probability of breakdown, p = once in 3 weeks i.e $\frac{1}{3}$

number of weeks n = 21

now,

mean, λ = np

= $\frac{1}{3}\times21$

= 7

P(X > 4) = 1 - ( P(X ≤ 4))

using Poisson distribution

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

Thus,

P(X = 0) = $\frac{e^{-7}7^0}{0!}$

= 0.00091

P(X = 1) = $\frac{e^{-7}7^1}{1!}$

= 0.00638

P(X = 2) = $\frac{e^{-7}7^2}{2!}$

= 0.02234

P(X = 3) = $\frac{e^{-7}7^3}{3!}$

= 0.05213

P(X = 4) = $\frac{e^{-7}7^4}{4!}$

= 0.09123

Thus,

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.00091 + 0.00638 + 0.02234 + 0.05213 + 0.09123

= 0.17299

Therefore,

P(X > 4) = 1 - 0.17299

= 0.827

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A lighthouse is located on an island 33 miles from the closest point on a straight shoreline. If the lighthouse light rotates cl

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Question:

A lighthouse is located on an island 3 miles from the closest point on a straight shoreline. If the lighthouse light rotates clockwise at a constant rate of 9 revolutions per minute, how fast does the beam of light move towards the point on the shore closest to the island when it is 52 miles from that point

Answer:

The beam of light moves at $16278\pi$ miles/min

Step-by-step explanation:

This question is illustrated with the attached image

Taking the instructions in the question, one at a time.

A revolution of 9 per minute implies that:

$\frac{d\theta}{dt} = \frac{9 * 2\pi\ rad}{1\ min}$

$\frac{d\theta}{dt} = 18\pi \frac{rad}{min}$

Take tan of the angle in the attachment:

$tan(\theta) =\frac{52}{3}$

Differentiate both sides with respect to time

$\frac{d\ tan(\theta)}{dt} =\frac{52}{3} * \frac{dx}{dt}$

Rewrite as:

$\frac{d\ tan(\theta)}{d\theta} * \frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

In calculus:

$sec^2(\theta) =\frac{d\ tan(\theta)}{d\theta}$ -- Chain rule

So:

$sec^2(\theta) *\frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

In trigonometry:

$sec^2(\theta) = tan^2(\theta) + 1$

So:

$(tan^2(\theta) + 1)\frac{d\theta}{dt} =\frac{52}{3} * \frac{dx}{dt}$

Recall that:

$tan(\theta) =\frac{x}{3}$

$((\frac{52}{3})^2 + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{52^2}{9} + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2704}{9} + 1)\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2704+9}{9})\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

$(\frac{2713}{9})\frac{d\theta}{dt} =\frac{1}{3} * \frac{dx}{dt}$

Recall that: $\frac{d\theta}{dt} = 18\pi \frac{rad}{min}$

$(\frac{2713}{9}) * 18\pi =\frac{1}{3} * \frac{dx}{dt}$

$2713 * 2\pi =\frac{1}{3} * \frac{dx}{dt}$

Multiply both sides by 3

$3 * 2713 * 2\pi =\frac{1}{3} * \frac{dx}{dt} * 3$

$3 * 2713 * 2\pi =\frac{dx}{dt}$

$16278\pi =\frac{dx}{dt}$

$\frac{dx}{dt} = 16278\pi$ miles/min

Hence:

The beam of light moves at $16278\pi$ miles/min

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Answer:

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Step-by-step explanation:

Plug each number in the formula

Do 16.5/100 which is 0.165

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To make it easier Use PEMDAS or BODMAS

You want to start with the numbers that are inside the parenthesis Multiplication comes first before addition.

So 0.165 x 15 is 2.475

Add the 1:

1+2.475= 3.475

Finally,

43,200(3.475) or 43,200 x 3.475

43,200 x 3.475= 150,120

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