Answer:
Hence the adjusted R-squared value for this model is 0.7205.
Step-by-step explanation:
Given n= sample size=20
Total Sum of square (SST) =1000
Model sum of square(SSR) =750
Residual Sum of Square (SSE)=250
The value of R ^2 for this model is,
R^2 = \frac{SSR}{SST}
R^2 = 750/1000 =0.75
Adjusted
:
Where k= number of regressors in the model.

double bracket factorisation
2 × 15 = 30
find two factors of 30 that add to give -7
(-10, +3)
2x^2 -10x + 3x - 15
2x(x-5) + 3 (x-5)
<em><u>(</u></em><em><u>2x</u></em><em><u>+</u></em><em><u>3</u></em><em><u>)</u></em><em><u> </u></em><em><u>(</u></em><em><u>x-5</u></em><em><u>)</u></em>
<em><u>THE</u></em><em><u> </u></em><em><u>ANSWER</u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>X-5</u></em>
The Bernoulli equation is almost identical to the standard linear ODE.

Compare to the basic linear ODE,

Meanwhile, the Riccati equation takes the form

which in special cases is of Bernoulli type if

, and linear if

. But in general each type takes a different method to solve. From now on, I'll abbreviate the coefficient functions as

for brevity.
For Bernoulli equations, the standard approach is to write


and substitute

. This makes

, so the ODE is rewritten as

and the equation is now linear in

.
The Riccati equation, on the other hand, requires a different substitution. Set

, so that

. Then you have



Next, setting

, so that

, allows you to write this as a linear second-order equation. You have



where

and

.
Answer:
-5/2
Step-by-step explanation:
With Distributive property, we get 4x + 4 = -6. Subtract 4 from both sides and we get 4x = -10. Divide both sides by 4 and we get x = -5/2.