X cannot equal +2, and can't be -1. If you factor it, then you notice you get 2(x+2)(x-2). If you use either one for the x, one will be zero and the won't, and vise-versa.
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.
Change 15% into a decimal
15% = 15/100 = 0.15
multiply 0.15 and 205,000 together
205000 x 0.15 = <span>30750
$30,750 is your answer
hope this helps</span>
Answer:
a) 0.125
b) 7
c) 0.875 hr
d) 1 hr
e) 0.875
Step-by-step explanation:l
Given:
Arrival rate, λ = 7
Service rate, μ = 8
a) probability that no requests for assistance are in the system (system is idle).
Let's first find p.
a) ρ = λ/μ

Probability that the system is idle =
1 - p
= 1 - 0.875
=0.125
probability that no requests for assistance are in the system is 0.125
b) average number of requests that will be waiting for service will be given as:
λ/(μ - λ)
= 7
(c) Average time in minutes before service
= λ/[μ(μ - λ)]
= 0.875 hour
(d) average time at the reference desk in minutes.
Average time in the system js given as: 1/(μ - λ)

= 1 hour
(e) Probability that a new arrival has to wait for service will be:
λ/μ =
= 0.875