The HA Theorem is a special case of the AAS postulate.
The following are the ages of 13 history teachers in a school district. 24, 27, 29, 29, 35, 39, 43, 45, 46, 49, 51, 51, 56 Notic
pishuonlain [190]
The five-number summary and the interquartile range for the data set are given as follows:
- Interquartile range: 50 - 29 = 21.
<h3>What are the median and the quartiles of a data-set?</h3>
- The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.
- The first quartile is the median of the first half of the data-set.
- The third quartile is the median of the second half of the data-set.
- The interquartile range is the difference between the third quartile and the first quartile.
In this problem, we have that:
- The minimum value is the smallest value, of 24.
- The maximum value is the smallest value, of 56.
- Since the data-set has odd cardinality, the median is the middle element, that is, the 7th element, as (13 + 1)/2 = 7, hence the median is of 43.
- The first quartile is the median of the six elements of the first half, that is, the mean of the third and fourth elements, mean of 29 and 29, hence 29.
- The third quartile is the median of the six elements of the second half, that is, the mean of the third and fourth elements of the second half, mean of 49 and 51, hence 50.
- The interquartile range is of 50 - 29 = 21.
More can be learned about five number summaries at brainly.com/question/17110151
#SPJ1
You can compute both the mean and second moment directly using the density function; in this case, it's
Then the mean (first moment) is
![E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x\,\mathrm dx=710](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7B80%7D%5Cint_%7B670%7D%5E%7B750%7Dx%5C%2C%5Cmathrm%20dx%3D710)
and the second moment is
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x^2\,\mathrm dx=\frac{1,513,900}3](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7B80%7D%5Cint_%7B670%7D%5E%7B750%7Dx%5E2%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7B1%2C513%2C900%7D3)
The second moment is useful in finding the variance, which is given by
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2=\dfrac{1,513,900}3-710^2=\dfrac{1600}3](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2%3D%5Cdfrac%7B1%2C513%2C900%7D3-710%5E2%3D%5Cdfrac%7B1600%7D3)
You get the standard deviation by taking the square root of the variance, and so
![\sqrt{V[X]}=\sqrt{\dfrac{1600}3}\approx23.09](https://tex.z-dn.net/?f=%5Csqrt%7BV%5BX%5D%7D%3D%5Csqrt%7B%5Cdfrac%7B1600%7D3%7D%5Capprox23.09)
I have no idea what the answer to that is or how to solve it but i LOVE UR PROFILE PICTURE
Answer:
Step-by-step explanation:
-5(10x^2 - 9)
