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mr Goodwill [35]
3 years ago
8

A tire manufacturer estimates that their tires last, an average of 40,000 miles, with a standard deviation of 5000. What is the

probability that a randomly chosen tire from this manufacturer lasts between 30,000 miles and 50,000 miles? (Give your answer as a decimal rounded to 4 decimal places.) check Reference Answer: 2. The daily high temperature on October 31 in a certain city is normally distributed with 50 and 0 8. What is the probability that the high temperature on October 31 in a randomly chosen year will be between 46 degrees and 58 degrees? (Give your answer as a decimal rounded to 4 decimal places.)
Mathematics
1 answer:
Oxana [17]3 years ago
3 0

Answer:

0.9554,0.8188

Step-by-step explanation:

Given that a tire manufacturer estimates that their tires last, an average of 40,000 miles, with a standard deviation of 5000.

X - duration of tires is N(40000,5000)

Or Z = \frac{x-40000}{5000} is N(0,1)

the probability that a randomly chosen tire from this manufacturer lasts between 30,000 miles and 50,000 miles

=P(30000

------------------------------------

2) Given that the daily high temperature on October 31 in a certain city is normally distributed with 50 and 8

Y - daily high temp on Oct 31 is N(50,8)

Or Z = \frac{x-50}{8} is N(0,1)

the probability that a randomly chosen tire from this manufacturer lasts between 46 degrees and 58 degrees

=P(46

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DENIUS [597]
3x^2 - 8x + 2 = 0\\ \\a=3 , \ \ b=-8 , \ \ c=2 \\ \\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a} =\frac{8-\sqrt{ (-8)^2-4 \cdot 3\cdot 2}}{2 \cdot 3} =\frac{8-\sqrt{ 64-24 }}{6} =\\ \\ =\frac{8-\sqrt{40 }}{6} = \frac{ 8-\sqrt{4\cdot 10 } }{6} = \frac{ 8-2\sqrt{ 10 }}{6} = \frac{2 (4-\sqrt{ 10 })}{6} = \frac{ 4-\sqrt{ 10 } }{3}

x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a} =\frac{8+\sqrt{ (-8)^2-4 \cdot 3\cdot 2}}{2 \cdot 3} = \frac{ 4+\sqrt{ 10 } }{3}



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