Answer:
y= (3/2)x-3
Step-by-step explanation:
We need two points to find the equation of a line. Let's take (2,0) and (4, 3).
In the equation y=mx+b, m represents the slope. To find the slope, we can calculate the change in y/change in x. For (2,0) and (4,3), the change in y is 3-0=3 and the change in x is 4-2=2. Therefore, our slope is 3/2.
Then, in the equation y=mx+b, we can plug 3/2 in for m to get y = (3/2)x+b. To find b, we can plug one point in, such as (2.0), to get 0=(3/2)(2) + b, 0=3+b, and b=-3, making our equation
y= (3/2)x-3
Answer: Honestly middle school grades, especially electives don't matter much. don't worry about it.
The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.
In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.
So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

Similarly,

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

Answer:
vertical angles?
Step-by-step explanation: