|2x + 6| - 4 = 20
First, add 4 to both sides. / Your problem should look like: |2x + 6| = 20 + 4
Second, simplify 20 + 4 to 24. / Your problem should look like: |2x + 6| = 24
Third, break down the problem into these 2 equations. / 2x + 6 = 24 and -(2x + 6) = 24
Fourth, solve the 1st equation: 2x + 6 = 24
Subtract 6 from both sides. / Your problem should look like: 2x = 24 - 6
Simplify 24 - 6 to 18. / Your problem should look like: 2x = 18
Divide both sides by 2. / Your problem should look like: x =

Simplify

to 9 / Your problem should look like:
x = 9
Fifth, solve the 2nd equation: -(2x + 6) = 24
Simplify brackets. / Your problem should look like: -2x - 6 = 24
Add 6 to both sides. / Your problem should look like: -2x = 24 + 6
Simplify 24 + 6 to 30. / Your problem should look like: -2x = 30
Divide both sides by -2. / Your problem should look like: x =

Simplify

to

/ Your problem should look like: x =

Simplify

to 15. / Your problem should look like:
x = -15
Sixth, collect all of your solutions. / Your problem should look like: x = -15, 9
Answer:
x = -15, 9 (C)
Answer: 0.0475
Step-by-step explanation:
Let x = random variable that represents the number of a particular type of bacteria in samples of 1 milliliter (ml) of drinking water, such that X is normally distributed.
Given: 
The probability that a given 1-ml will contain more than 100 bacteria will be:
![P(X>100)=P(\dfrac{X-\mu}{\sigma}>\dfrac{100-85}{9})\\\\=P(Z>1.67)\ \ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Zz)=1-P(Z](https://tex.z-dn.net/?f=P%28X%3E100%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B100-85%7D%7B9%7D%29%5C%5C%5C%5C%3DP%28Z%3E1.67%29%5C%20%5C%20%5C%20%5C%20%5BZ%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C1.67%29%5C%20%5C%20%5C%20%5BP%28Z%3Ez%29%3D1-P%28Z%3Cz%29%5D%5C%5C%5C%5C%3D1-%200.9525%3D0.0475)
∴The probability that a given 1-ml will contain more than 100 bacteria
0.0475.
Might be A, I don’t know.