Mr. Harrison grades his test on a scale of 1 to 10. The dot plots below show the grades for his first test and the last test of the year. Pic below What is the difference between the means of the scores for the two tests?

Work Shown:

F = first test data set

F = {1,3,5,5,7,7,7,8,8,8,8,9,9,9,9,9,9,9,9,10,10,10,10,10,10}

m = mean of set F

m = (sum of values in set F)/(number of values in set F)

m = (1+3+5+5+7+7+7+8+8+8+8+9+9+9+9+9+9+9+9+10+10+10+10+10+10)/25

m = 199/25

m = 7.96

L = last test data set

L = {3,4,4,5,5,6,6,6,7,7,7,7,7,7,8,8,8,8,8,9,9,9,9,10,10}

n = mean of set L

n = (sum of values in set L)/(number of values in set L)

n = (3+4+4+5+5+6+6+6+7+7+7+7+7+7+8+8+8+8+8+9+9+9+9+10+10)/25

n = 177/25

n = 7.08

d = difference in means

d = m - n

d = 7.96 - 7.08

d = 0.88

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 hope this is what you were looking for.

see attached graph