A soft drink machine outputs a mean of 27 ounces per cup. The machine's output is normally distributed with a standard deviation
1 answer:
Answer: 0.6065
Step-by-step explanation:
Given : The machine's output is normally distributed with
Let x be the random variable that represents the output of machine .
z-score :
For x= 21 ounces
For x= 28 ounces
Using the standard normal distribution table , we have
The p-value :
Hence, the probability of filling a cup between 21 and 28 ounces= 0.6065
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Answer:
B
Step-by-step explanation:
Given
f(x) =
The denominator cannot be zero as this would make f(x) undefined.
Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.
x + 2 = 0 ⇒ x = - 2
The equation of the vertical asymptote is x = - 2
Let x be the number of bags of Nature's Recipe Venison Meal & Rice Canine 20 percent protein, y be the number of bags of Nutro Max Natural Dog Food 27 percent protein and z be the number of bags of PetSmart Premier Oven Baked Lamb 25 percent protein a dog breeder buys. He wants to make 300 pounds of a mix. Then
20x+17.5y+30z=300.
Now
- in 20x pounds of Nature's Recipe Venison Meal & Rice Canine 20 percent protein is 0.2·20x=4x pounds of protein;
- in 17.5y pounds of Nutro Max Natural Dog Food 27 percent protein is 0.27·17.5y=4.725y pounds of protein;
- in 30z pounds of PetSmart Premier Oven Baked Lamb 25 percent protein is 0.25·30z=7.5z pounds of protein;
- in 300 pounds of mix containing 22 percent protein is 0.22·300=66 pounds of protein.
Then 4x+4.725y+7.5z=66.
You get a system
From the first equation Substitute it into the second equation:
Simplify it:
y must be divided by 20, then
- y=0, x=9, z=10-6=4;
- y=20, x=16, z is not a whole number;
- y=40, x=23, z is not a whole number;
- y=60, x=30, z=10-20-35<0;
- thus, all other possible z will be <0.
Answer: 9 bags with 1st mix, 0 bags with 2nd mix and 4 bags with 3rd mix.