b) The university gym opens at 9 am on Saturday mornings. One Saturday morning at 8:55 am there are 15 students outside the gym waiting for it to open. Should you use the same approach from part (a) to calculate the expected number of smokers among these 15 students

The complete question is:

At a university, 13% of students smoke.

a) Calculate the expected number of smokers in a random sample of 100 students from this university:

b) The university gym opens at 9 am on Saturday mornings. One Saturday morning at 8:55 am there are 27 students outside the gym waiting for it to open. Should you use the same approach from part (a) to calculate the expected number of smokers among these 27 students?

Part a is easy, because is a random sample, we can expect that just 13% of these 100 students to be smokers, and 13% of 100 is 13, so we can expect 13 of those 100 students to be smokers.

b) This time we do not have a random sample, our sample is a sample of 15 students who go to the gym in the early morning, so our sample is biased. (And we do not know if this bias is related to smoking or not, and how that relationship is), so we can't use the same approach that we used in the previous part.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752