The slope-intercept form:

m - slope
b - y-intercept → (0, b)
We have the point (0, 5). Therefore b = 5.
We have: y = mx + 5
Other point (3, 2). Substitute the coordinates of the point to the equation of the line:
<em>subtract 5 from both sides</em>
<em>divide both sides by 3</em>

<h3>Answer: f(x) = -1x + 5 = -x + 5</h3>
Answer:
Step-by-step explanation:
y = (x-1)²
vertex (1,0)
y-intercept = (0-1)² = 1
Answer:
-2y + 6 = -12
Step-by-step explanation:
well the answer is 9, y=9 :)
PLEASE MARK ME BRAINLIEST!!!
Answer:
The probability of the system being down in the next hour of operation is 0.3.
Step-by-step explanation:
We have a transition matrix from one period to the next (one hour) that can be written as:
![T=\left[\begin{array}{ccc}&R&D\\R&0.7&0.3\\D&0.2&0.8\end{array}\right]](https://tex.z-dn.net/?f=T%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%26R%26D%5C%5CR%260.7%260.3%5C%5CD%260.2%260.8%5Cend%7Barray%7D%5Cright%5D)
We can represent the state that system is initially running with the vector:
![S_0=\left[\begin{array}{cc}1&0\end{array}\right]](https://tex.z-dn.net/?f=S_0%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5Cend%7Barray%7D%5Cright%5D)
The probabilties of the states in the next period can be calculated using the matrix product of the actual state and the transition matrix:

That is:
![S_1=S_0\cdot T= \left[\begin{array}{cc}1&0\end{array}\right]\cdot \left[\begin{array}{cc}0.7&0.3\\0.2&0.8\end{array}\right]= \left[\begin{array}{cc}0.7&0.3\end{array}\right]](https://tex.z-dn.net/?f=S_1%3DS_0%5Ccdot%20T%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.7%260.3%5C%5C0.2%260.8%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.7%260.3%5Cend%7Barray%7D%5Cright%5D)
With the inital state as running, we have a probabilty of 0.7 that the system will be running in the next hour and a probability of 0.3 that it will be down.