Any student who answers yes to one or more of these questions will be considered to have been involved in some type of cheating. an important part of your study is to determine if bayview's proportion of cheating students, = 0.4111, is significantly less (in the statistical sense) than the article's proportion, p0 = 0.56. while lower, the sample proportion could be within the range of what you would expect due to random variation in samples taken from a population where p = 0.56. you were able to obtain 90 responses to your survey with a sample proportion of = 0.4111. in this case, the hypothesis test you established in question 1 generated a p-value of 0.0022 leading to a rejection of h0 at = 0.05. what would the p-value have been if only 35 students had responded to your survey? what decision would you have made about your hypothesis test in that situation? assume the same sample proportion, = 0.4111. a. 0.0668, fail to reject h0 b. 0.0576, reject h0 c. 0.038, reject h0

answer: the answer is 127 degrees

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:

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