DRUGS R US is a large manufacturer of various kinds of liquid vitamins. The quality control department has noted that the bottles of vitamins marked 6 ounces vary in content with a standard deviation of 0.4. Assume the contents of the bottles are normally distributed. 4.23% of bottles will have more than how many ounces?

4.32% of the bottles will have more than 12.86 ounces.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 6 ounces

Standard deviation = σ = 0.4 ounce

Let the z-score of 4.32% of the bottles be z' and the corresponding ounces be x'

The z-score for any is the value minus the mean then divided by the standard deviation.

z' = (x' - μ)/σ

P(x > x') = P(z > z') = 0.0432

Using the normal distribution table

P(z > z') = 1 - P(z ≤ z') = 0.0432

P(z ≤ z') = 1 - 0.0432 = 0.9568

z' = 1.715

1.715 = (x' - 6)/0.4

x' - 6 = 4(1.715)

x' = 6 + 6.86 = 12.86 ounces.

4.32% of the bottles will have more than 12.86 ounces.

Hope this Helps!!!

12 servings. 4 times 3= 12

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2