2(12+18) is not equivalent to 24+18
To find the z-score for a weight of 196 oz., use

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
-3 is the slope, use y2 - y1 over x2 - x1 equation to solve
Answer:
10 segments
Step-by-step explanation:
AB, BC, CD, DE, AC, AD, AE, BD, BE, CE
Answer:
the correct answer is a
Step-by-step explanation: