|x-6| < 1 + 10 (move 10 to the right side)
|x-6| < 11
1)x-6 < 11 (here x >= 6)
2)-x+6 < 11 (here x < 6)
1)x<17
2)x>-5
so we get -5 < x < 17
the answer: (-5; 17)
Answer:
You must survey 784 air passengers.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
The margin of error is:
95% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
Assume that nothing is known about the percentage of passengers who prefer aisle seats.
This means that 
, which is when the largest sample size will be needed.
Within 3.5 percentage points of the true population percentage.
We have to find n for which M = 0.035. So
You must survey 784 air passengers.
Answer:
.65% / hr = .0065 / hr
2.75 days * 24 hrs / day = 66 hrs
66 hrs * .0065 / hr = .429
So our amount will become
A = A0 ( 1 + .429) = 9700 * 1.429 = 13860
Answer:
Least number is 60.
Step-by-step explanation:
Let the number of CDs has Jo be represented by x + 10. Where x = 0, 1, 2, 3....
Then, since Ken has twice as many CDs as Jo, the number of CDs that Ken has can be expressed as;
2(x + 10) = 2x + 20
Maisie has three times as many CDs as Ken, then the number of CDs that Maisie has can be expressed as;
3(2x + 20) = 6x + 60
Thus to determine the least number of CDs that Maisie can have, let x = 0.
3(2x + 20) = 6x + 60
= 6(0) + 60
= 60
Therefore, the least number of CDs that Maisie can have is 60.
