You can see how this works by thinking through what's going on.
In the first year the population declines by 3%. So the population at the end of the first year is the starting population (1200) minus the decline: 1200 minus 3% of 1200. 3% of 1200 is the same as .03 * 1200. So the population at the end of the first year is 1200 - .03 * 1200. That can be written as 1200 * (1 - .03), or 1200 * 0.97
What about the second year? The population starts at 1200 * 0.97. It declines by 3% again. But 3% of what??? The decline is based on the population at the beginning of the year, NOT based no the original population. So the decline in the second year is 0.03 * (1200 * 0.97). And just as in the first year, the population at the end of the second year is the population at the beginning of the second year minus the decline in the second year. So that's 1200 * 0.97 - 0.03 * (1200 * 0.97), which is equal to 1200 * 0.97 (1 - 0.03) = 1200 * 0.97 * 0.97 = 1200 * 0.972.
So there's a pattern. If you worked out the third year, you'd see that the population ends up as 1200 * 0.973, and it would keep going like that.
So the population after x years is 1200 * 0.97x
Answer:
i think if it were me i whould divide 300 dide 15
Step-by-step explanation:
Answer should be 108 would you mind sending a visual so I can be sure
Answer:
Approximately, the 90% confidence interval for the students' mean IQ score is between 129.045 - 130.956
Step-by-step explanation:
The formula to use to solve this question is called the Confidence Interval formula.
Confidence interval =
x ± z × ( σ/ (√n) )
Where:
x = the sample mean = 130
z = the z-value for 90% confidence = 1.645
σ = standard deviation = 7
n = sample size = 145
130 ± 1.645 × (7/√145)
130 ± 0.9562687005
130 - 0.9562687005 = 129.0437313
130 + 0.9562687005 = 130.9562687005
Therefore, approximately, the 90% confidence interval for the students' mean IQ score is between 129.045 - 130.956
