Answer:
The percentage of the bag that should have popped 96 kernels or more is 2.1%.
Step-by-step explanation:
The random variable <em>X</em> can be defined as the number of popcorn kernels that popped out of a mini bag.
The mean is, <em>μ</em> = 72 and the standard deviation is, <em>σ</em> = 12.
Assume that the population of the number of popcorn kernels that popped out of a mini bag follows a Normal distribution.
Compute the probability that a bag popped 96 kernels or more as follows:
Apply continuity correction:


*Use a <em>z</em>-table.
The probability that a bag popped 96 kernels or more is 0.021.
The percentage is, 0.021 × 100 = 2.1%.
Thus, the percentage of the bag that should have popped 96 kernels or more is 2.1%.
Answer:
I mean how can I answer this???????????????????????
Answer:
I think it is 192
Step-by-step explanation:
Volume of cube: 5 x 4 x 8 = 160
Area of diagonal: 160/2 = 80
Length of diagonal: 80 / 0.5 x 3 = 480
AB = 480 / 5 x 2 = 192
volume = 4/3 x PI x r^3
r = 10/2 =5
V = 4/3 x 3.14 x 5^3 = 523.333
523.3 cubic inches