Answer:
115 mg
Step-by-step explanation:
The 50th percentile has a corresponding z-score of z = 0. The z-score, for any given value of caffeine, X, is given by:
Therefore, if z=0, then X must be equal to the mean value of the normal distribution, which is 115 mg. Thus, the 50th percentile value of caffeine content is 115 mg.
The coeficient is 2.6 so
2.6x+23.4
if you want to factor out the coeficient then divide the whole thing by 2.6
(2.6x+23.4)/2.6=x+9
2(x+3) means 2 times (x+3)
so the factored out coeficient is
2.6(x+9)
The correct answer is G. s+7,653≥13,300 or s≥13,300-7653 => s≥5,647
Good luck!!!
The answer would be 152 with a remainder of 1