Answer:
3,498.921
Step-by-step explanation:
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
In the model
Log (salary) = B0 + B1LSAT +B2GPA +B3log(libvol) +B4log(cost)+B5 rank+u
The hypothesis that rank has no effect on log (salary) is H0:B5 = 0. The estimated equation (now with standard errors) is
Log (salary) = 8.34 + .0047 LSAT + .248 GPA + .095 log(libvol)
(0.53) (.0040) (.090) (.033)
+ .038 log(cost) – .0033 rank
(.032) (.0003)
n = 136, R2 = .842.
The t statistic on rank is –11(i.e. 0.0033/0.0003), which is very significant. If rank decreases by 10 (which is a move up for a law school), median starting salary is predicted to increase by about 3.3%.
(ii) LSAT is not statistically significant (t statistic ≈1.18) but GPA is very significance (t statistic ≈2.76). The test for joint significance is moot given that GPA is so significant, but for completeness the F statistic is about 9.95 (with 2 and 130 df) and p-value ≈.0001.
Answer:
8,000
Step-by-step explanation:
You can simplify it to (-4+24)^3a
Answer:
y^3/(27 x^3)
Step-by-step explanation:
Simplify the following:
((3 x)/y)^(-3)
((3 x)/y)^(-3) = (y/(3 x))^3:
(y/(3 x))^3
Multiply each exponent in y/(3 x) by 3:
(y^3)/((3 x)^3)
Multiply each exponent in 3 x by 3:
y^3/(3^3 x^3)
3^3 = 3×3^2:
y^3/(3×3^2 x^3)
3^2 = 9:
y^3/(3×9 x^3)
3×9 = 27:
Answer: y^3/(27 x^3)
