The amount of mustard dispensed from a standard dispenser is normally distributed with a mean of 0.9oz and a variance of 0.01oz. If a family of six buys ten hotdogs total and dispenses mustard onto each one using the standard dispenser, what is the probability that more than 3 of them will have more than one ounce of mustard?

the probability that more than 3 of them will have more than one ounce of mustard is 0.05975

Step-by-step explanation:

Given te data in the question;

let x rep the quantity of mustard on a hotdog ( in oz)

so

X → N ( u = 0.9, α² = 0.01 )

now P( more than one ounces of mustard on a hotdog) will be

P( X>1 ) = P( Z > ((1-0.9)/(√0.01)) ) = P( Z > 1 ) = 1 - P( Z < 1 )

= 1 - 0.84134 = 0.15866

Also let Y rep the number of hotdogs that have more than one ounces of mustard, so

Y → Bin( n = 10, p = 0.15866 )

P( y = y ) = (¹⁰ _y) ( 0.15855)⁰ ( 1 - 0.15866 )^10-y, y = 0,1,2,3, 10

so required probability = P(Y > 3) = 1 - P(Y ≤ 3)

= 1 - P(Y=0) - P(Y=1) - P(Y=2) - P(Y=3)

=1 -  (¹⁰ ₀) ( 0.15855)⁰ (1 - 0.15855)¹⁰⁻⁰ - (¹⁰ ₁) ( 0.15855)¹ (1 - 0.15855)¹⁰⁻¹ - (¹⁰ ₂) ( 0.15855)² (1 - 0.15855)¹⁰⁻² - (¹⁰ ₃) ( 0.15855)³ (1 - 0.15855)¹⁰⁻³

= 0.05975

Therefore the probability that more than 3 of them will have more than one ounce of mustard is 0.05975

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