Answered: ch.8 question #12 Express the…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section5.6: Exponential And Logarithmic Equations

Problem 6E

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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?
Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?

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ch.8 question #12 Express the confidence interval 0.837±0.033 in open interval form (i.e., (0.155,0.855)). ____________________