Answer:
i dont know sup man give me your girl
Step-by-step explanation:
The pattern here is the number times 3 then plus 1
We can check this by plugging it in.
1×3=3
3+1=4
4 is the next number in our sequence so the pattern works.
We can continue to check this with the rest of our sequence.
4×3=12
12+1=13
13 is the next number in our sequence so the pattern works.
13×3=39
39+1=40
40 is the next number in our sequence so the pattern works.
40×3=120
120+1=121
121 is the next number in our sequence so the pattern works.
We can find the next numbed in the sequence by continuing the patter
121×3=363
363+1=364
So the next number in the sequence is 364
Answer:
○ The graph of y = -2x² + 3 opens downward and is shifted up.
Step-by-step explanation:
According to the Quadratic Equation, <em>y = Ax² + Bx + C</em>, <em>C</em><em> </em>acts<em> </em>like a y-intercept, and in this case, since both graphs shift up [because both are positive values], we do not pay attention to those. Now, we come over to our <em>A</em>, which makes a BIG difference because both graphs open in opposite directions [one negative, one positive]. With this being stated, we have our answer.
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** Extended information on Parabolas
Opens down → -A
Opens up → A
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.
