These ad agencies must focus on their target audience, which are the students. Hence, they should gather data on the pool that will surely comprise of students. For agency B, social media posting is not a good source pool. It's true that students are very participative and opinionated in social media. However, they can't be sure that these are students. Some parents are in social media, as well. Some are working individuals, and some are out of school youth. Unlike agency A, agency B has to sort out profiles first and identify which ones are students. Hence, agency A will produce a fair sample of the student population because it is unarguably true that everyone in the school enrollment data are students.
The answer is B.
An equation is y = 2x-1
And then draw a line or something to show that the equation is true
Answer:
The upper limit of a 95% confidence interval for the population mean would equal 83.805.
Step-by-step explanation:
The standard deviation is the square root of the variance. Since the variance is 25, the sample's standard deviation is 5.
We have the sample standard deviation, not the population, so we use the t-distribution to solve this question.
T interval:
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 15 - 1 = 14
Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of 0.95(
). So we have T = 1.761
The margin of error is:
M = T*s = 1.761*5 = 8.805.
The upper end of the interval is the sample mean added to M. So it is 75 + 8.805 = 83.805.
The upper limit of a 95% confidence interval for the population mean would equal 83.805.
2|x-3|+1=7
2|x-3|=6
|x-3|=3
x-3 = 3 or x-3 = -3
x=0 or 6
Answer:
equation is
2(x+10x)=1540
width= 70ft
length= 700ft
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