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
Answer is 10(2k+5)
-----------------------------------
Work Shown:
20k+50 = 10*2k + 10*5
20k+50 = 10(2k+5)
Note how we can distribute the 10 back in to check our work
10(2k+5) = 10(2k)+10(5) = 20k+50
so that confirms we have the right answer
Another thing to notice is that 10 is the largest factor that we can pull out of 20k and 50. The value 10 is the GCF (greatest common factor) of 20 and 50.
Answer:
<em>No, he should have set the sum of ∠AED and ∠DEC equal to 180°, rather then setting ∠AED and ∠DEC equal to each other</em>
Step-by-step explanation:
Find the diagram attached
If line AC and BD intersects, then m<AED + m<DEC = 180 (sum of angle on a straight line is 180 degrees)
Given
m<AED = 16x+8
m<DEC = 76 degrees
16x + 8 + 76 = 180
16x + 84 = 180
16x = 180-84
16x = 96
x = 96/16
x = 6
Hence the value of x is 6
Hence the correct option is <em>No, he should have set the sum of ∠AED and ∠DEC equal to 180°, rather then setting ∠AED and ∠DEC equal to each other</em>