His/her lowered score was most likely due to statistical regression.
<h3>How to determine the reason?</h3>
The missing options in the question are:
A. compensation rivalry B. Demoralization C. Differential selection
D. Testing E. Statistical regression
From the question, we have:
- September = 99th percentile
- February = 90th percentile
A change (whether higher or lower) in the score is caused by statistical regression.
This is so because several variables could attribute to the change in the score.
The relationship between these variables is referred to as statistical regression
Read more about statistical regression at:
brainly.com/question/25987747
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Answer:
The percentage rate per annum is 2.8%
Step-by-step explanation:
We need to multiply the principal by the time, then divide the interest by the product of P×T, so:
1,600 × 5 = 8,000
224 ÷ 8,000 = 0.028
Then we convert our final product to a percentage.
0.028 to a percent is 2.8%
Answer: Area of ΔABC is 2.25x the area of ΔDEF.
Step-by-step explanation: Because equilateral triangle has 3 equal sides, area is calculated as
![A=\frac{\sqrt{3} }{4} a^{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B4%7D%20a%5E%7B2%7D)
with a as side of the triangle.
Triangle ABC is 20% bigger than the original, which means its side (a₁) measures, compared to the original:
a₁ = 1.2a
Then, its area is
![A_{1}=\frac{\sqrt{3} }{4}(1.2a)^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%3D%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B4%7D%281.2a%29%5E%7B2%7D)
![A_{1}=\frac{\sqrt{3} }{4}1.44a^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%3D%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B4%7D1.44a%5E%7B2%7D)
Triangle DEF is 20% smaller than the original, which means its side is:
a₂ = 0.8a
So, area is
![A_{2}=\frac{\sqrt{3} }{4} (0.8a)^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%3D%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B4%7D%20%280.8a%29%5E%7B2%7D)
![A_{2}=\frac{\sqrt{3} }{4} 0.64a^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%3D%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B4%7D%200.64a%5E%7B2%7D)
Now, comparing areas:
![\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )](https://tex.z-dn.net/?f=%5Cfrac%7BA_%7B1%7D%7D%7BA_%7B2%7D%7D%3D%20%28%5Cfrac%7B%5Csqrt%7B3%7D.1.44a%5E%7B2%7D%20%7D%7B4%7D%29%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D.0.64a%5E%7B2%7D%20%7D%20%29)
2.25
<u>The area of ΔABC is </u><u>2.25x</u><u> greater than the area of ΔDEF.</u>
Answer:
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used 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 = 14 - 1 = 13
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of
. So we have T = 3.0123
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 664.14 - 224.85 = $439.29
The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).