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.
Answer:
-1, 2,5,8
Step-by-step explanation:
nth term =a+(n-1) *d
-1+(n-1) *3
-1+3n-3
3n-4
t2 =3*2-4
6-4
2
t3=3*3-4
9-4
5
t4=3*4-4
12-4
8
56.4 *10 = 564
564/10 =
282/5 <span>in simplest form</span>

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