Of a random sample of 200 auditors, 105 indicated some measure of agreement with this statement: "cash flow is an important indicator of profitability." [assume the sample is large](a)test at the 10% significance level against a two-sided alternative the null hypothesis that one-half of the members of this population would agree with this statement. (b)find and interpret the p-value of this test.

a)At the 10% significance level, we fail to reject the null hypothesis.

b) P-value is ≈ 0.48

Step-by-step explanation:

Let p be the proportion of the auditors agree with the statement  "Cash flow is an important indicator of profitability."

Then,

: p=0.5

: p≠0.5

Test statistic can be calculated as:

z= where

Using these numbers we get:

z=≈0,707

and p(z) is ≈ 0.48. Since 0.48>0.10, the result is not significant at 90% level. Therefore we fail to reject the null hypothesis.

english, cause i can’t you

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.