Answer:
x = 3
x = (-1)/2
x = 13/4
Step-by-step explanation:
Solve for x:
(2 x)/3 + 15 = 17
Put each term in (2 x)/3 + 15 over the common denominator 3: (2 x)/3 + 15 = (2 x)/3 + 45/3:
(2 x)/3 + 45/3 = 17
(2 x)/3 + 45/3 = (2 x + 45)/3:
1/3 (2 x + 45) = 17
Multiply both sides of (2 x + 45)/3 = 17 by 3:
(3 (2 x + 45))/3 = 3×17
(3 (2 x + 45))/3 = 3/3×(2 x + 45) = 2 x + 45:
2 x + 45 = 3×17
3×17 = 51:
2 x + 45 = 51
Subtract 45 from both sides:
2 x + (45 - 45) = 51 - 45
45 - 45 = 0:
2 x = 51 - 45
51 - 45 = 6:
2 x = 6
Divide both sides of 2 x = 6 by 2:
(2 x)/2 = 6/2
2/2 = 1:
x = 6/2
The gcd of 6 and 2 is 2, so 6/2 = (2×3)/(2×1) = 2/2×3 = 3:
Answer: x = 3
______________________________________________________
Solve for x:
3 x - x + 8 = 7
Grouping like terms, 3 x - x + 8 = (3 x - x) + 8:
(3 x - x) + 8 = 7
3 x - x = 2 x:
2 x + 8 = 7
Subtract 8 from both sides:
2 x + (8 - 8) = 7 - 8
8 - 8 = 0:
2 x = 7 - 8
7 - 8 = -1:
2 x = -1
Divide both sides of 2 x = -1 by 2:
(2 x)/2 = (-1)/2
2/2 = 1:
Answer: x = (-1)/2
_______________________________________
Solve for x:
4 (2 x - 6) = 2
Divide both sides of 4 (2 x - 6) = 2 by 4:
(4 (2 x - 6))/4 = 2/4
4/4 = 1:
2 x - 6 = 2/4
The gcd of 2 and 4 is 2, so 2/4 = (2×1)/(2×2) = 2/2×1/2 = 1/2:
2 x - 6 = 1/2
Add 6 to both sides:
2 x + (6 - 6) = 1/2 + 6
6 - 6 = 0:
2 x = 1/2 + 6
Put 1/2 + 6 over the common denominator 2. 1/2 + 6 = 1/2 + (2×6)/2:
2 x = 1/2 + (2×6)/2
2×6 = 12:
2 x = 1/2 + 12/2
1/2 + 12/2 = (1 + 12)/2:
2 x = (1 + 12)/2
1 + 12 = 13:
2 x = 13/2
Divide both sides by 2:
x = (13/2)/2
2×2 = 4:
Answer: x = 13/4
Answer:
it is likely that a female will be picked
Step-by-step explanation:
the probability of selecting a female = 21/25....or 84%
You're trying to find constants
![a_0,a_1,a_2](https://tex.z-dn.net/?f=a_0%2Ca_1%2Ca_2)
such that
![\hat y=a_0+a_1\hat x+a_2{\hat x}^2](https://tex.z-dn.net/?f=%5Chat%20y%3Da_0%2Ba_1%5Chat%20x%2Ba_2%7B%5Chat%20x%7D%5E2)
. Equivalently, you're looking for the least-square solution to the following matrix equation.
![\underbrace{\begin{bmatrix}1&6&6^2\\1&3&3^2\\\vdots&\vdots&\vdots\\1&9&9^2\end{bmatrix}}_{\mathbf A}\underbrace{\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}}_{\mathbf x}=\underbrace{\begin{bmatrix}100\\110\\\vdots\\70\end{bmatrix}}_{\mathbf b}](https://tex.z-dn.net/?f=%5Cunderbrace%7B%5Cbegin%7Bbmatrix%7D1%266%266%5E2%5C%5C1%263%263%5E2%5C%5C%5Cvdots%26%5Cvdots%26%5Cvdots%5C%5C1%269%269%5E2%5Cend%7Bbmatrix%7D%7D_%7B%5Cmathbf%20A%7D%5Cunderbrace%7B%5Cbegin%7Bbmatrix%7Da_0%5C%5Ca_1%5C%5Ca_2%5Cend%7Bbmatrix%7D%7D_%7B%5Cmathbf%20x%7D%3D%5Cunderbrace%7B%5Cbegin%7Bbmatrix%7D100%5C%5C110%5C%5C%5Cvdots%5C%5C70%5Cend%7Bbmatrix%7D%7D_%7B%5Cmathbf%20b%7D)
To solve
![\mathbf{Ax}=\mathbf b](https://tex.z-dn.net/?f=%5Cmathbf%7BAx%7D%3D%5Cmathbf%20b)
, multiply both sides by the transpose of
![\mathbf A](https://tex.z-dn.net/?f=%5Cmathbf%20A)
, which introduces an invertible square matrix on the LHS.
![\mathbf{Ax}=\mathbf b\implies\mathbf A^\top\mathbf{Ax}=\mathbf A^\top\mathbf b\implies\mathbf x=(\mathbf A^\top\mathbf A)^{-1}\mathbf A^\top\mathbf b](https://tex.z-dn.net/?f=%5Cmathbf%7BAx%7D%3D%5Cmathbf%20b%5Cimplies%5Cmathbf%20A%5E%5Ctop%5Cmathbf%7BAx%7D%3D%5Cmathbf%20A%5E%5Ctop%5Cmathbf%20b%5Cimplies%5Cmathbf%20x%3D%28%5Cmathbf%20A%5E%5Ctop%5Cmathbf%20A%29%5E%7B-1%7D%5Cmathbf%20A%5E%5Ctop%5Cmathbf%20b)
Computing this, you'd find that
![\mathbf x=\begin{bmatrix}a_0\\a_1\\a_2\end{bmatrix}\approx\begin{bmatrix}121.119\\-3.786\\-0.175\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20x%3D%5Cbegin%7Bbmatrix%7Da_0%5C%5Ca_1%5C%5Ca_2%5Cend%7Bbmatrix%7D%5Capprox%5Cbegin%7Bbmatrix%7D121.119%5C%5C-3.786%5C%5C-0.175%5Cend%7Bbmatrix%7D)
which means the first choice is correct.
I NEED BOINTS IM
SORRY GOOD LORD BUTTTTTTT MAYBE C