Answer:
Subtract 5 from each side
Step-by-step explanation:
This would put all coefficients on one side.
Answer:
7x=−x+24 7 x = − x + 24
Step-by-step explanation:
The equations we solved in the last section simplified nicely so that we could use the. Our strategy will involve choosing one side of the equation to be the variable side, and step by step, to isolate the variable terms on one side of the equation
Answer:
Part A:
The probability that all of the balls selected are white:

Part B:
The conditional probability that the die landed on 3 if all the balls selected are white:

Step-by-step explanation:
A is the event all balls are white.
D_i is the dice outcome.
Sine the die is fair:
for i∈{1,2,3,4,5,6}
In case of 10 black and 5 white balls:






Part A:
The probability that all of the balls selected are white:


Part B:
The conditional probability that the die landed on 3 if all the balls selected are white:
We have to find 
The data required is calculated above:

-8
On the number line you either start at -3 or -5 and you got your answer