Answer:
(c) 118/V100 11.8
Step-by-step explanation:
We use the central limit theorem to solve this question.
The Central Limit Theorem estabilishes that, for a random variable X, with mean 
and standard deviation 
, a large sample size can be approximated to a normal distribution with mean and standard deviation 
In this problem, we have that:
So
So the correct answer is:
(c) 118/V100 11.8
<h3>Answer:</h3>
Yes, ΔPʹQʹRʹ is a reflection of ΔPQR over the x-axis
<h3>Explanation:</h3>
The problem statement tells you the transformation is ...
... (x, y) → (x, -y)
Consider the two points (0, 1) and (0, -1). These points are chosen for your consideration because their y-coordinates have opposite signs—just like the points of the transformation above. They are equidistant from the x-axis, one above, and one below. Each is a <em>reflection</em> of the other across the x-axis.
Along with translation and rotation, <em>reflection</em> is a transformation that <em>does not change any distance or angle measures</em>. (That is why these transformations are all called "rigid" transformations: the size and shape of the transformed object do not change.)
An object that has the same length and angle measures before and after transformation <em>is congruent</em> to its transformed self.
So, ... ∆P'Q'R' is a reflection of ∆PQR over the x-axis, and is congruent to ∆PQR.
Answer: A) .1587
Step-by-step explanation:
Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.
i.e. 
and 
Let x denotes the amount of soda in any can.
Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.
Then, the probability that a randomly selected can will need to go through the mentioned process = probability that a randomly selected can has more than 12.50 ounces of soda poured into it =
![P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587](https://tex.z-dn.net/?f=P%28x%3E12.50%29%3D1-P%28x%5Cleq12.50%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B12.50-12.30%7D%7B0.20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1%29%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.8413%5C%20%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%5C%5C%5C%3D0.1587)
Hence, the required probability= A) 0.1587
