D. Ledger lines are used to extend musical staff. I think my answer is correct because ledger lines represent notes that surpass the bottom and up limits.
I've heard better reviews for UCLA. (I think the social life is better, or so im told )
The standard error of the difference of sample means is 0.444
From the complete question, we have the following parameters
<u>Canadians</u>
- Sample size = 50
- Mean = 4.6
- Standard deviation = 2.9
<u>Americans</u>
- Sample size = 60
- Mean = 5.2
- Standard deviation = 1.3
The standard error of a sample is the quotient of the standard deviation and the square root of the sample size.
This is represented as:
![SE = \frac{\sigma}{\sqrt n}](https://tex.z-dn.net/?f=SE%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%20n%7D)
The standard error of the Canadian sample is:
![SE_1 = \frac{2.9}{\sqrt{50}}](https://tex.z-dn.net/?f=SE_1%20%3D%20%5Cfrac%7B2.9%7D%7B%5Csqrt%7B50%7D%7D)
So, we have:
![SE_1 = 0.41](https://tex.z-dn.net/?f=SE_1%20%3D%200.41)
The standard error of the American sample is:
![SE_2 = \frac{1.3}{\sqrt{60}}](https://tex.z-dn.net/?f=SE_2%20%3D%20%5Cfrac%7B1.3%7D%7B%5Csqrt%7B60%7D%7D)
So, we have:
![SE_2 = 0.17](https://tex.z-dn.net/?f=SE_2%20%3D%200.17)
The standard error of the difference of sample means is then calculated as:
![SE= \sqrt{SE_1^2 + SE_2^2}](https://tex.z-dn.net/?f=SE%3D%20%5Csqrt%7BSE_1%5E2%20%2B%20SE_2%5E2%7D)
This gives
![SE= \sqrt{0.41^2 + 0.17^2}](https://tex.z-dn.net/?f=SE%3D%20%5Csqrt%7B0.41%5E2%20%2B%200.17%5E2%7D)
![SE= \sqrt{0.197}](https://tex.z-dn.net/?f=SE%3D%20%5Csqrt%7B0.197%7D)
Take square roots
![SE= 0.444](https://tex.z-dn.net/?f=SE%3D%200.444)
Hence, the standard error of the difference of sample means is 0.444
Read more about standard errors at:
brainly.com/question/6851971