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:
The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y with 0 and solve for x. x= i, −i, 3,−2
Step-by-step explanation:
Answer:
1.) n-10
2.) 4 / x=x (dont trust me on this)
3.) -8n+1
4.) 1/2 - 19 ( dont trust me on this)
5.)3/5-21 or 3/5>21
6.)x^+15
Step-by-step explanation:
I tried double check it
its been awhile
X ⇒ ln (2)
e^3 x - 8/e^2 x - 4
Suppose that the farmer had bought the rice at x dollars per bag and had sold them at a 25% markup. How much did the bags cost him before he added the markup? 1.25x =$75 results in $75/1.25, or $60 per bag.
If he sold 25 bags, his profit would be 1.25($60/bag)(25 bags) = $1875.
I very seriously doubt that the rice was $7500 per bag. Perhaps you meant $75/bag...?