The answer is 114.5%.
Let the original population be denoted by x.
Now, let's go through the percentage increase per year.
<u>Year 1</u>
<u>Year 2</u>
- 1.2x (1 + 25%)
- 1.2x (1.25)
- 1.5x
<u>Year 3</u>
- 1.5x (1 + 30%)
- 1.5x (1.3)
- 1.95x
<u>Year 4</u>
- 1.95x (1 + 10%)
- 1.95x (1.1)
- 2.145x
Overall increase : 214.5% - 100% = 114.5%
Hence, the overall percentage increase in these 4 years is 114.5%.
Answer:
a) 17.09 hours
b) The 95% confidence interval estimate of the population mean flying time for the Pilots is between 31.91 hours and 66.09 hours
Step-by-step explanation:
We have the standard deviation of the sample, so we use the t distribution to solve this question.
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 = 49 - 1 = 48
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 48 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.0106
The margin of error is:
M = T*s = 2.0106*8.5 = 17.09
s is the standard deviation of the sample. 17.09 hours is the answer for a.
The lower end of the interval is the sample mean subtracted by M. So it is 49 - 17.09 = 31.91 hours
The upper end of the interval is the sample mean added to M. So it is 49 + 17.09 = 66.09 hours
The 95% confidence interval estimate of the population mean flying time for the Pilots is between 31.91 hours and 66.09 hours
Answer: The unit for Warren’s data is the number of cars sold each day.
Sorry for the answer being so late!
Answer:
k = - 8
Step-by-step explanation:
Subtract 9 from both sides to get the constants on one side and the variables on the other
9 - 2k = 25
- 2k = 25 - 9
Simplify
- 2k = 16
Divide by -2 on both sides to isolate the variable, then simplify once more.
- 2k/ -2 = 16/ -2
k = 16/ -2
k = -8