Margin of error, e = Z*SD/Sqrt (N), where N = Sample population
Assuming a 95% confidence interval and substituting all the values;
At 95% confidence, Z = 1.96
Therefore,
0.23 = 1.96*1.9/Sqrt (N)
Sqrt (N) = 1.96*1.9/0.23
N = (1.96*1.9/0.23)^2 = 262.16 ≈ 263
Minimum sample size required is 263 students.
One of the ways to find the answer is to find how much 1 percent of 150 is. To do that we just have to divide 150 by 100:
150 / 100 = 1.5
Now we can divide 9 by 1.5 to know how much percent of 150 it is:
9 / 1.5 = 6
Let's check it:
150 * 6% = 150 * 6/100 = 15 * 6/10 = 3 * 6/2 = 18/2 = 9
It's correct then :)
2x+1=13 x=6 so an ok answer would be: 2x+10=40 x=15
Answer:
c
Step-by-step explanation:
took the test
