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kiruha [24]
3 years ago
10

Reserve Problems Chapter 4 Section 5 Problem 2 A laptop company claims up to 10.1 hours of wireless web usage for its newest lap

top battery life. However, reviews on this laptop shows many complaints about low battery life. A survey on battery life reported by customers shows that it follows a normal distribution with mean 9.5 hours and standard deviation 30 minutes. (a) What is the probability that the battery life is at least 10.1 hours? Round your answer to four decimal places (e.g. 98.7654).
Mathematics
1 answer:
Viktor [21]3 years ago
4 0

Answer:

0.1151

Step-by-step explanation:

Given that a laptop company claims up to 10.1 hours of wireless web usage for its newest laptop battery life. However, reviews on this laptop shows many complaints about low battery life. A survey on battery life reported by customers shows that it follows a normal distribution with mean 9.5 hours and standard deviation 30 minutes.

If X is the battery life then X is N(9.5, 0.5)

(we convert into one unit for calculations purpose here hours)

a) the probability that the battery life is at least 10.1 hours

=P(X\geq 10.1)\\=P(Z\geq \frac{10.1-9.5}{0.5} )\\= P(Z\geq 1.2)\\=0.1151

we get the probability that the battery life is at least 10.1 hours is 0.1151

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Answer:

x = 6

Step-by-step explanation:

3 (2x - 5) = 21

=> 6x - 15 = 21

=> 6x = 21 + 15

=> 6x = 36

=> x = 36 / 6

=> x = 6

OR

3 (2x - 5) = 21

=> 2x - 5 = 21 / 3

=> 2x - 5 = 7

=> 2x = 7 + 5

=> 2x = 12

=> x = 12 / 2

=> x = 6

Hope it helps :)

Please mark my answer as the brainliest

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3 years ago
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Nuetrik [128]

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Step-by-step explanation:

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3 years ago
A lighthouse is located on a small island 2 km away from the nearest point P on a straight shoreline and its light makes four re
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Answer:

62.8 km per min

Step-by-step explanation:

Let the diagram of this situation is shown below,

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Also, 1 revolution = 2\pi radians

So, the change in angle with respect to t ( time ),

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Let x be the distance of beam from P,

\implies \tan \theta = \frac{x}{2}

Differentiating with respect to x,

\sec^2 \theta \frac{d\theta}{dt}=\frac{1}{2}\frac{dx}{dt}

(1+\tan^2 \theta) (8\pi) = \frac{1}{2}\frac{dx}{dt}

(1+\frac{x^2}{2^2})(8\pi) = \frac{1}{2}\frac{dx}{dt}

If x = 1 km,

(1+\frac{1}{4})8\pi =\frac{1}{2}\frac{dx}{dt}

\frac{5}{4}8\pi = \frac{1}{2}\frac{dx}{dt}

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see the picture.

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