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
solution
1 inch = 2.54 centimeters
Therefore, to convert to cubic measurements
(1 in)³ = (2.54 cm)³
Thus;
1 in³ = 16.387 cm³
Hence 305 cubic inches will be equivalent to
305 × 16.387
= 4998.035 cm³
206.047
two hundred and six, forty seven thousandths
200 + 6 + 0.04 + 0.007
Answer:
48.48%
Step-by-step explanation:
Let's assume that there is a number N of women.
32% of these are smokers, then there are 0.32*N smokers
then 68% of these are non-smokers, then there are 0.68*N non-smokers.
Let's assume that the probability of having a ectopic pregnancy for a non-smoker is p (and the probability for a smoker will be 2*p)
Then the number of women with an ectopic pregnancy that are non-smokers is:
p*0.68*N
The number of women with an ectopic pregnancy that are smokers is:
2*p*0.32*N
Then the total number of women with an ectopic pregnancy will be:
p*0.68*N + 2*p*0.32*N
The percentage of women having an ectopic pregnancy that are smokers is equal to the quotient between the number of women with an ectopic pregnancy that are smokers and the total number of women with an ectopic pregnancy, all that times 100%.
The percentage is:

Taking p and N as common factors, we get:

Then we get:

Answer:
x=2/3
Step-by-step explanation:
-3x+4=2
-3x=-2
x=2/3