appendixa_s - salary = exp(10.6 ≈ $40,134.84 When exper =...

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APPENDIX A SOLUTIONS TO PROBLEMS A.1 (i) $566. (ii) The two middle numbers are 480 and 530; when these are averaged, we obtain 505, or $505. (iii) 5.66 and 5.05, respectively. (iv) The average increases to $586 while the median is unchanged ($505). A.3 If price = 15 and income = 200, quantity = 120 – 9.8(15) + .03(200) = –21, which is nonsense. This shows that linear demand functions generally cannot describe demand over a wide range of prices and income. A.5 The majority shareholder is referring to the percentage point increase in the stock return, while the CEO is referring to the change relative to the initial return of 15%. To be precise, the shareholder should specifically refer to a 3 percentage point increase. A.7 (i) When exper = 0, log( salary ) = 10.6; therefore,
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Unformatted text preview: salary = exp(10.6) ≈ $40,134.84. When exper = 5, salary = exp[10.6 + .027(5)] ≈ $45,935.80. (ii) The approximate proportionate increase is .027(5) = .135, so the approximate percentage change is 13.5%. (iii) 100[(45,935.80 – 40,134.84)/40,134.84) ≈ 14.5%, so the exact percentage increase is about one percentage point higher. A.9 (i) The relationship between yield and fertilizer is graphed below. 117 fertilizer 50 100 120 121 yield 122 (ii) Compared with a linear function, the function yield = .120 + .19 fertilizer has a diminishing effect, and the slope approaches zero as fertilizer gets large. The initial pound of fertilizer has the largest effect, and each additional pound has an effect smaller than the previous pound. 118...
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This note was uploaded on 05/03/2009 for the course ECON 418 taught by Professor Breman during the Spring '08 term at Arizona.

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appendixa_s - salary = exp(10.6 ≈ $40,134.84 When exper =...

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