(2, ∞) ..................................
Answer:
68+72+85=225
plus 79 =304 divided by 4 = 76
plus 89 =314 divided by 4 = 78.5
plus 91=316 divided by 4 = 79
so the answer is C
plz brainlest
It would be 5.75 divided by 5, making the answer 1.15.
<span>2x(x - 3)(x + 2)
= </span><span>2x(x^2 - 3x + 2x - 6)
</span>= 2x(x^2 - x - 6)
= 2x^3 - 2x^2 - 12x
Answer:
Given :The monthly demand for a product is normally distributed with mean = 700 and standard deviation = 200.
To Find :
1. What is probability demand will exceed 900 units in a month?
2. What is probability demand will be less than 392 units in a month?
Solution:
![\mu = 700 \\\sigma = 200](https://tex.z-dn.net/?f=%5Cmu%20%3D%20700%20%5C%5C%5Csigma%20%3D%20200)
We are supposed to find probability demand will exceed 900 units in a month.
Formula : ![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
We are supposed to find P(Z>900)
Substitute x = 900
![z=\frac{900-700}{200}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B900-700%7D%7B200%7D)
![z=1](https://tex.z-dn.net/?f=z%3D1)
Refer the z table.
P(Z<900)=0.8413
P(Z>900)=1-P{(Z<900)=1-0.8413=0.1587
So, the probability that demand will exceed 900 units in a month is 0.1587.
Now we are supposed to find probability demand will be less than 392 units in a month
We are supposed to find P(Z<392)
Substitute x = 392
![z=\frac{392-700}{200}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B392-700%7D%7B200%7D)
![z=-1.54](https://tex.z-dn.net/?f=z%3D-1.54)
refer the z table
P(Z<900)=0.0618
So, probability that demand will be less than 392 units in a month is 0.0618.