Given: The burning time of a very large candle is normally distributed with mean $(\mu)$ of 2500 hours and standard deviation $(\sigma)$ of 20 hours.

Let X be a random variable that represent the burning time of a very large candle.

Formula: $Z=\dfrac{X-\mu}{\sigma}$

For X = 2470

$Z=\dfrac{2470-2500}{20}\\\\\Rightarrow\ Z=\dfrac{-30}{20}\\\\\Rightarrow\ Z=\dfrac{-3}{2}\\\\\Rightarrow\ Z=-1.5$

So, the z-score they corresponds to a lifespan of 2470 hours. =-1.5

Hence, the correct option is C.-1.5.

7 0

2 years ago

Find the value of a and b when x=12

garri49 [273]

A= 360
B= 2(12)²(7)/120= 2016/120=16.8

5 0

3 years ago

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The price p in dollars and the quantity x sold of a certain product obey the demand equation -1/5x+100 (0≤ x ≤ 500)

olga nikolaevna [1]

Answer:

See below

Step-by-step explanation:

We have:

p = -1/5x + 100 = -0.2x + 100

a) The revenue is:

R = px
R = x( -0.2x + 100)
R= -0.2x² + 100x

b) x = 200, find R:

R = -0.2(200²) + 100(200) = 12000

c) This is a quadratic function and the maximum value is obtained at vertex.

x = -100/(-0.2*2) = 250 is the required quantity

d) The max revenue is obtained when -0.2x + 100 at max:

The maximum possible is p = 100 when x = 0

5 0

3 years ago

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