The following r commands and outputs show scores on the first quiz (maximum score of 10 points) for eighth-grade students in an introductory level french course. the instructor grouped the student in the course as follows: - group 1: never studied foreign language before but have good english skills - group 2: never studied foreign language before and have poor english skills - group 3: studied at least one other foreign language quiz scores group 1 group 2 group 3 5 10 mean standard deviation 8 u = 6 s1 = 2 u2 = 3 $2 = 2.8 v3 = 8 $3 = 2.6 > test < -aov(scores group, data=quiz) > summary(test) df sum sq mean sq f value pr(> f) group 2 30 2.5 0.177 residuals 5 30 6 15 use r commands and outputs above to answer part a. through h. assume that quiz scores are normally distributed.

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: