Answer:
1.199 ≤ x ≤ 1.201
Step-by-step explanation:
You want ...
|y -7| ≤ 0.005
Substituting the expression for y, you have ...
|5x+1 -7| ≤ 0.005
|x -1.2| ≤ 0.001 . . . . . simplify, divide by 5
-0.0001 ≤ x -1.2 ≤ 0.001 . . . . . "unfold" the absolute value
1.199 ≤ x ≤ 1.201 . . . . . . . . . . . .add 1.2
Oh ok fine I just want to make sure you know that I don’t want you to the route cover 4/5
6y^2 + 6y + 2y + 2 = 6y^2 + 8y + 2
The way you do this is basically multiplying the values in the brackets with 6y, and then adding that to the same values multiplied by 2.
You have, on the bottom the number of cars and on the left the corresponding probability.
The probability of having 4 cars is 0.05; the probability of having 5 cars is 0.05 again.
Therefore, the probability of having more than 3 cars is P(4) + P(5) = 0.10 which corresponds to 10%.
The correct answer is 0.10
PEMDAS
(Parenthesis, Exponents, Multiplication and Division, Adding and Subtracting)