A researcher was testing the number of popcorn kernels that popped out of a mini bag of 100 kernels after being cooked for the specific time in the microwave. the researcher found that on average 72 currently parked with a standard deviation of 12.what percentage of the bag should have popped 96 kernels or more?

The percentage of the bag that should have popped 96 kernels or more is 2.1%.

Step-by-step explanation:

The random variable X can be defined as the number of popcorn kernels that popped out of a mini bag.

The mean is, μ = 72 and the standard deviation is, σ = 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 z-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%.

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

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step-by-step explanation: