The box plots show the weights, in pounds, of the dogs in two different animal shelters.

Weights of Dogs in Shelter A

2 box plots. The number line goes from 6 to 30. For the weights of dogs in shelter A, the whiskers range from 8 to 30, and the box ranges from 17 to 28. A line divides the box at 21. For shelter B, the whiskers range from 10 to 28, and the box ranges from 16 to 20. A line divides the box at 18.

Weights of Dogs in Shelter B

Which is true of the data in the box plots? Select three choices.

The median weight for shelter A is greater than that for shelter B.

The median weight for shelter B is greater than that for shelter A.

The data for shelter A are a symmetric data set.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

step-by-step explanation:

d would equal 12