7 1/3 ÷ 4 Please help
2 answers:
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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%.
Answer:
Step-by-step explanation:
Given
Paper = 20 slips
Word: PENNSYLVANIA
Required
Determine P(Multiple of 4 and V)
The sample size of the 20 slips is:
The outcomes of multiples of 4 is:
So, the probability of multiples of 4 is:
The sample size of PENNSYLVANIA is:
The outcome of V is:
So, the probability of V is:
So, the required probability is: P(Multiple of 4 and V)
Express as percentage