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Answer:

0.0000

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Step-by-step explanation:

Given that a tobacco company claims that the amount of nicotine in its cigarettes is a random variable with mean 2.2 mg and standard deviation .3 mg.

i.e. population parameters are

$\mu =2.2 \\s = 0.3$

The approximate probability that the sample meanwould have been as high or higher than 3.1

$=P(X\geq 3.1)\\=P(Z\geq \frac{3.1-2.2}{\frac{0.3}{\sqrt{100} } } )\\=P(Z\geq 30)\\$

=0.0000

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Answer:

x=70°

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x+110=180 linear pair

x=180-110=70

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