An 27. Match each of the normal curves to its mean u and standarddeviation o.a. μ = -2, o = 2c. μ = 4, o = 1i.ii.iii.iv.b. μ = -2, o = 4d. μ = 4, o = 3-15 -10 -5 0 5 10 15-15 -10. -5 0 5 10 15-15 -10 -5 0 51015-15 -10 -5 0 5 10 15

Normal distribution:The normal distribution is a continuous probability distribution that is…

Answered: Match each of the normal curves to its…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.6: Summarizing Categorical Data

Problem 31PPS

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Planetary Velocity The following table gives the mean velocity of planets in their orbits versus their mean distance from the sun. Note that 1AU astronomical unit is the mean distance from Earth to the sun, abut 93 million miles. Planet d=distance AU v=velocity km/sec Mercury 0.39 47.4 Venus 0.72 35.0 Earth 1.00 29.8 Mars 1.52 24.1 Jupiter 5.20 13.1 Saturn 9.58 9.7 Uranus 19.20 6.8 Neptune 30.05 5.4 Astronomers tell us that it is reasonable to model these data with a power function. a Use power regression to express velocity as a power function of distance from the sun. b Plot the data along with the regression equation. c An asteroid orbits at a mean distance of 3AU from the sun. According to the power model you found in part a, what is the mean orbital velocity of the asteroid?
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II. Direction: Sketch and identify the areas of the regions under the normal curve corresponding to the following standard normal values or z – values: 1. to the left of z = 0 2. to the left of z = -1.14 3. to the left of z = 2.3 4. between z = -1 and z = 3 5. between z = -3 and z = 3
Given: xy=15. Find the equation of the normal line at x=5 a. 3x+5y=30 b. 5x-3y=16 c. 3x=5y d. 5x+3y=34
Answer d, e, & f.
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Match each of the normal curves to its mean and standard deviation o. a. μ = -2, o = 2 c. μ = 4,o=1 i. ii. F iv. -15 -10 -5 b. μ = -2,0 = 4 d. μ = 4, o = 3 -15 -10 T 0 5 10 T T -15 -10 -5 0 5 10 15 15 -15 -10 -5 0 5 10 15 -5 0 5 10 15