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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%.
Step-by-step explanation:
We have,
Mass of ball is 9 kg
Initial speed, u = 5 m/s
Final speed, v = -2 m/s (negative as it bounces off)
(a) The change in velocity of the bowling ball is :
(b) Change of momentum of the ball is :
|p| = 630 kg-m/s
(c) Impulse momentum theorem states that the change in momentum of the ball is equal to the impulse exerted on the ball. So, impulse is 630 kg-m/s.
(d) Impulse is also given by :