What was the age distribution of nurses in Great Britain at the time of Florence Nightingale? Suppose we have the following information. Note: In 1851 there were 25,466 nurses in Great Britain. Age range (yr) 20-29 30-39 40-49 50-59 60-69 70-79 80+ Midpoint x 24.5 34.5 44.5 54.5 64.5 74.5 84.5 Percent of nurses 5.7% 9.7% 19.5% 29.2% 25.0% 9.1% 1.8% (a) Using the age midpoints x and the percent of nurses, do we have a valid probability distribution? Explain. (b) Use a histogram to graph the probability distribution in part (a). (c) Find the probability that a British nurse selected at random in 1851 would be 60 years of age or older. (d) Compute the expected age μ of a British nurse contemporary to Florence Nightingale. (e) Compute the standard deviation σ for ages of nurses shown in the distribution.

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

i believe the answer is the polygon, heptagon.

there are no answer choices

step-by-step explanation: