Tutoring Services: The Community College Survey of Student Engagement reports that 46% of the students surveyed rarely or never use peer or other tutoring resources. Suppose that in reality 40% of community college students never use tutoring services available at their college. In a simulation we select random samples from a population in which 40% do not use tutoring. For each sample we calculate the proportion who do not use tutoring. If we randomly sample 500 students at a time, what will be the mean and standard error of the sampling distribution of sample proportions

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.

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step-by-step explanation:

what do you mean by "representation"? graph? table? etc.

a) 40 more; b) old factory:   10%, new factory:   16%; c) week 4

explanation:

a) at week 0, we see in the graph that the new factory produces 190 specialty items.  

using the function, substituting 0 in for w,

this is a difference of 230-190 = 40.

b) the function for the old factory is written in the form f(x) = a(1+r)ˣ, where a is the original amount, x is the amount of time and r is the growth rate.   the equation for the old factory is ; 230 is in place of a and is the original amount; w is the amount of time; which means 1+r = 1.1.   this means that r must be 0.1 or 10%.

for the new factory, we find the percent of increase to find the growth rate; the formula for this is

between the first two points, the amount of change is 220-190 = 30; this gives us

30/190 = 0.158 ≈ 0.16 = 16%

c) at week 0, the old factory has 230 while the new factory has 190.

at week 1, the old factory has 230(1.1) = 253 while the new factory has 220.

at week 2, the old factory has 230(1.1²) = 278.3 while the new factory has 252.

at week 3, the old factory has 230(1.1³) = 306.13 while the new factory has 290.

at week 4, the old factory has 230(1.1⁴) = 336.743 while the new factory has 337.   this is the first week that the new factory surpasses the old.

the correct answer is c) 2.