A. Mr. Kent interviewed the 54 students as they are going to leave the school, it is not considered to be a random sample. It is because a random sample is when a set is taken from a population. Mr. Kent interviewed the 54 who are going to leave, meaning, he didn't take a set out of that 54, he took all of them. So it is not a random sample.
b. The question that Mr. Kent asked is considered to be a leading question, so it does not seem biased.
c. If there are 54 respondents.
51 = yes, the rest is no.
= 54 - 51 = 3
= 3 is now divided to 54 = 3/54
= giving an answer of 0.0555
= 0.0555 x 100
= 5.6%
= The percent of responses that says 'no' is 5.6%
The Answer is 32%.
Because for two workers for 40 hours each, it will cost 640
and when you do .32*2000 it equals 640 also
Hope I helped :)<span />
You're given two points for f(t) and g(t) each. Two points, (x1,y1), and (x2,y2). In this case, x1 is always x1=0, since the time t starts counting at the earliest year given.
We know the form y=mx+b. m can be found a la (y2-y1)/(x2-x1), but remember x1 is 0. b can be found at f(0) or g(0), which we're already given; y1 for both cases.
I'm sorry, but I'm not giving a direct answer. Knowing the method, and being careful with calculations, should be more than sufficient.<span>it would be 22395-20808/5=317.4 f(t)
26508-26155/5= 70.6
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If it takes 20 minute to get to the airport, and he wants to be there an hour and 5 minutes before his flight leaves, we have to subtract the time he leaves (9:50) by 1 hour and 5.
9:50 - 1 hour and 5 minutes.
9:50 - 65 minutes.
9:50 - 65 = 8:45.
Which means he must be at the airport at 8:45.
BUT,
The question is asking for the latest time that Quincy can LEAVE for the airport, which means it's 8:45 - 20 minutes, since it takes 20 minutes to get to the airport:
8:45 - 20 = 8:25