On a true/false test, amy answered the first three questions wrong but answered the rest of the questions correctly. on the same test, scott answered exactly two questions wrong. he answered the last question wrong, the 26 questions before it correctly, and the question before that wrong. if amy and scott were both incorrect on the same question exactly once, what is the greatest possible percent of the total number of questions that amy could have answered correctly?

89.29%

Step-by-step explanation:

Given that Scott answered exactly two questions wrong. He answered the last question wrong, the 26 questions before it correctly, and the question before that wrong.  Since Amy and Scott were both incorrect on the same question exactly once, Scott would have done wrong only in I or II or III question.  If I question was wrong by both, total questions = 28, if II wrong total questions =29 and for iii total questions =30

Greatest possible percent would occur if total questions are less

i.e. when total =28

Greatest percent =

the thermometer is precise but not accurate

step-by-step explanation:

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!