The solution of x is x = 0, x =2 and x = -5
<h3>How to solve for the unknown?</h3>
The equation is given as:
2x^3+6x^2-20x=0
Factor out x
x(2x^2+6x-20)=0
Split the equation
x = 0 or 2x^2+6x - 20=0
Solve for x in 2x^2+6x - 20=0
Expand the equation
2x^2 + 10x - 4x - 20 = 0
Factorize the equation
(2x - 4)(x +5) = 0
Split the equation
2x - 4 = 0 and x + 5 = 0
Solve for x
x =2 and x = -5
Hence, the solution of x is x = 0, x =2 and x = -5
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Answer:
164m (to 3 s.f.)
Step-by-step explanation:
Please see the attached picture for the full solution.
Answer:
0.44
Step-by-step explanation:
Given the estimated logistic regression model on risk of having squamous cell carcinoma
-4.84 + 4.6*(SMOKER)
SMOKER = 0 (non-smoker) ; 1 (SMOKER)
What is the predicted probability of a smoker having squamous cell carcinoma?
exp(-4.84 + 4.6*(SMOKER)) / 1 + exp(-4.84 + 4.6*(SMOKER))
SMOKER = 1
exp(-4.84 + 4.6) / 1 + exp(-4.84 + 4.6)
exp^(-0.24) / (1 + exp^(-0.24))
0.7866278 / 1.7866278
= 0.4402863
= 0.44
Answer:
The determinant is 15.
Step-by-step explanation:
You need to calculate the determinant of the given matrix.
1. Subtract column 3 multiplied by 3 from column 1 (C1=C1−(3)C3):
![\left[\begin{array}{ccc}-25&-23&9\\0&3&1\\-5&5&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%26-23%269%5C%5C0%263%261%5C%5C-5%265%263%5Cend%7Barray%7D%5Cright%5D)
2. Subtract column 3 multiplied by 3 from column 2 (C2=C2−(3)C3):
![\left[\begin{array}{ccc}-25&-23&9\\0&0&1\\-5&-4&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%26-23%269%5C%5C0%260%261%5C%5C-5%26-4%263%5Cend%7Barray%7D%5Cright%5D)
3. Expand along the row 2: (See attached picture).
We get that the answer is 15. The determinant is 15.
