Answer:
Step-by-step explanation:
Solve the y
y -3 = 1/2(x + 2)
y = 1/2(x + 2) + 3 Add
Pick points that you wish to substitute for x values.
x = 0 x = 2 x = 4
y = 1/2(x + 2) + 3 y = 1/2(x + 2) + 3 y = 1/2(x + 2) + 3
y = 1/2(0 + 2) + 3 y = 1/2(2 + 2) + 3 y = 1/2(4 + 2) + 3
y = 1/2(2) + 3 y = 1/2(4) + 3 y = 1/2(6) + 3
y = 1 + 3 y = 2 + 3 y = 3 + 3
y = 4 y = 5 y = 6
Locate the following point on graph (x, y); (0, 4), (2, 5), (4, 6)
The image is kinda blurry, but from what I can gather, the notation in the attachment means
![p_X[x_k]\equiv\mathbb P(X=k)=\begin{cases}(1-p)^{k-1}p&\text{for }k=1,2,\ldots\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=p_X%5Bx_k%5D%5Cequiv%5Cmathbb%20P%28X%3Dk%29%3D%5Cbegin%7Bcases%7D%281-p%29%5E%7Bk-1%7Dp%26%5Ctext%7Bfor%20%7Dk%3D1%2C2%2C%5Cldots%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
Then
a)
![p_X[x_k\le4]\equiv\mathbb P(X\le4)=\mathbb P(X=1)+\mathbb P(X=2)+\mathbb P(X=3)+\mathbb P(X=4)](https://tex.z-dn.net/?f=p_X%5Bx_k%5Cle4%5D%5Cequiv%5Cmathbb%20P%28X%5Cle4%29%3D%5Cmathbb%20P%28X%3D1%29%2B%5Cmathbb%20P%28X%3D2%29%2B%5Cmathbb%20P%28X%3D3%29%2B%5Cmathbb%20P%28X%3D4%29)
b)

c)
Answer:
yall werid
Step-by-step explanation:
Answer:
The correct option is 1.
Step-by-step explanation:
If a quadratic equation is defined as

then the quadratic formula is

The given equation is

Here, a=1, b=-2 and c=-3.
Substitute these values in the above formula.


Therefore, the correct option is 1.