United Airlines' flights from Boston to Dallas are on time 80 % of the time. Suppose 7 flights are randomly selected, and the number on-time flights is recorded. 1.The probability that at least 3 flights are on time is

14.5

Step-by-step explanation:

Given data:

Success rate = 80% = 0.8

No of flights = 7

No of failures = 1 – 0.8 = 0.2

Solution

at least 3 flights are on time is

= 7C3 * (0.80)^3 * (0.80)^1  +  7C7 * (0.80)^7 * (0.80)^1

= 35 * 0.512 * 0.8 + 1 * 0.21 * 0.8

= 14.336 + 0.168

= 14.5

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

1,8,15,22,29,36,43,50

step-by-step explanation: