If the candidate only wants a 0.2% margin of error at a 95% confidence level, what size of sample is needed?

n = 22

Step-by-step explanation:

Using

n = p(1-p)(Zc/E)²

From table E = 0.2 while p= 0.37

So substituting

We have n= 22.4

Thats a sample size of approximately 22

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

answer: 33 ( approx)

step-by-step explanation:

since, in the linear regression formula, the slope is the a in the equation y = b + ax ( by comparing it with the general equation of line y = mx + c )

where,

here given table is,

price ($), x   :           18         23       25       29       30       33

pairs sold, y :         16         13         10       8         6       2

so, ,   ,   ,

and   ,

thus, slop of the given line is,

⇒  

⇒   ≈ 33