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.
Answer:
x^2 = 10^2 - 9^2
x^2 = 100 -81 = 19
x = 4.3588989435
The triangle sides are 10, 9 and 4.3588989435
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Answer:
they would have 1 scoop per cone and 80 kds
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they would get one fav
SOMINE PLSSS ANSWER THIS ONE
THE ANSWER IS C.
PLEASE VOTE BRAINLIEST
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