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Mrrafil [7]
3 years ago
11

A certain candle is designed to last nine hours. However, depending on the wind, air bubbles in the wax, the quality of the wax,

and the number of times the candle is re-lit, the actual burning time (in hours) is a uniform random variable with a = 5.5 and b = 9.5.
Suppose one of these candles is randomly selected.
(a) Find the probability that the candle burns at least six hours.
(b) Find the probability that the candle burns at most seven hours.
(c) Find the mean burning time. Find the probability that the burning time of a randomly selected candle will be within one standard deviation of the mean. (Round your answer to four decimal places.)
(d) Find a time t such that 25% of all candles burn longer than t hours.
Mathematics
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:

Step-by-step explanation:

a) P(X > 6) = (9.5-6)/(9.5-5.5) = 3.5/4 = 0.875

b) P(X < 7) = (7-5.5)/(9.5-5.5) = 1.5/4 = 0.375

c) E(X) = (9.5+5.5)/2 = 7.5

Standard deviation = (9.5-5.5)/sqrt(12) =4/3.46 = 1.154

P= 1.156*2/(9.5-5.5) = 2.308/4 = 0.577

d) P(X > t) = 0.25

 (9.5-t) /(9.5-5.5) = 0.3

9.5-t = 1.2

t = 9.5-1.2 = 8.3

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The perimeter of the area of the pen the farmer intends to build for his

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Response:

i) The length and width of the rectangular pen are; <em>x</em>, and \dfrac{100 - x}{2}, therefore;

  • The area is; A = \dfrac{1}{2} \cdot x \cdot (100 - x)

ii) \hspace{0.5 cm}\dfrac{dA}{dx}  = 50 - x

\dfrac{d^2A}{dx^2} =  -1

iii) The value of <em>x</em> that makes the area as large as possible is x = 50

<h3>How is the function for the area and the maximum area obtained?</h3>

Given:

The length of fencing the farmer has = 100 m

Part of the area of the pen is a permanent stone wall.

Let <em>x</em> represent the length of the stone wall, we have;

2 × Width = 100 m - x

Therefore;

Width, <em>w</em>, of the rectangular pen, w = \mathbf{\dfrac{100 - x}{2}}

Area of a rectangle = Length × Width

Area of the rectangular pen, is therefore;

  • A = x \times \dfrac{100 - x}{2} = \underline{\dfrac{1}{2} \cdot x \cdot (100 - x)}

ii) \hspace{0.5 cm} \mathbf{\dfrac{dA}{dx}}, and \mathbf{\dfrac{d^2A}{dx^2} } are found as follows;

\dfrac{dA}{dx} = \mathbf{\dfrac{d}{dx} \left(  \dfrac{1}{2} \cdot x \cdot (100 - x) \right)} = \underline{50 - x}

\dfrac{d^2A}{dx^2} = \mathbf{ \dfrac{d}{dx} \left( 50 - x\right)} = \underline{-1}

iii) The value of <em>x</em> that makes the area as large as possible is given as follows;

Given that the second derivative, \dfrac{d^2A}{dx^2} =-1, is negative, we have;

At the maximum area, \dfrac{dA}{dx} = \mathbf{0}, which gives;

\dfrac{dA}{dx} = 50 - x = 0

x = 50

  • The value of x that makes the area as large as possible is <em>x</em>  =<u> 50</u>

Learn more about the maximum value of a function here:

brainly.com/question/19021959

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