The answer is C. Good luck
Answer:
Step-by-step explanation:
Answer:
y-b/m = x
Step-by-step explanation:
y = mx + b
~Subtract b to both sides
y - b = mx
~Divide m to both sides
y-b/m = x
Best of Luck!
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:
![P( X\geq 96)=P( X>96+0.50)](https://tex.z-dn.net/?f=P%28%20X%5Cgeq%2096%29%3DP%28%20X%3E96%2B0.50%29)
![=P( X>96.50)\\=P(\frac{X-\mu}{\sigma}>\frac{96.50-72}{12})\\=P(Z>2.04)\\=1-P(Z](https://tex.z-dn.net/?f=%3DP%28%20X%3E96.50%29%5C%5C%3DP%28%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cfrac%7B96.50-72%7D%7B12%7D%29%5C%5C%3DP%28Z%3E2.04%29%5C%5C%3D1-P%28Z%3C2.04%29%5C%5C%3D1-0.97932%5C%5C%3D0.02068%5C%5C%5Capprox%200.021)
*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%.