Answer:
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Step-by-step explanation:
We have the standard deviation for 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 = 25 - 1 = 24
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 2.064
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 81.5 - 4.21 = 77.29
The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Answer: -26
Step-by-step explanation:
2(n - 4)= 3(n + 6)
2n -8 = 3n + 18
+8 +8
2n = 3n + 26
-3 -3
-1n = 26
n= -26
(x^2 +5x+2.5^2) +4+5
(x+2.5)^2+9
vertex -2.5,9
Answer:
Lead-the-market pay strategies. An employer may choose to establish an internal compensation strategy that is in excess of the pay rates in the prevailing marketplace. This compensation strategy may increase the supply of candidates, increase selection rates of qualified applicants, decrease employee turnover, increase morale and productivity, or prevent unionization efforts. However, prior to implementing a lead compensation strategy, an organization should carefully consider what benefits it expects to realize from such a strategy, keeping in mind that this type of structure has the greatest propensity of increasing overall labor costs.
Step-by-step explanation:
(32 ft) x (30 ft + 5/12 ft) = (960 ft² + 13-1/3 ft²) = 973-1/3 feet²
