A student earned grades of Upper AA, Upper DD, Upper AA, Upper CC, and Upper BB. Those courses had the corresponding numbers
1 answer:
Answer:
- The grade point average is 2.92
- The student didn't make the Dean's list because the student's GPA is less than 3.0
Step-by-step explanation:
- I take the grades as A,D,A,C,B not AA,DD,AA,CC,BB.
- I take numbers of credit hours as 4,2,2,3,1 not as 44, 22, 22, 33, and 11.
Since quality points to letter grades are A=4; B=3; C=2; D=1; F=0, weighted mean is the sum of the qulity points times corresponding credit hours divided by the total credit hours:
≈ 2.92
Since 2.92<3.0, the student is not in Dean's list.
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Hey there, hope I can help!
Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
Lets use the quadratic formula now
Multiply the numbers 2 * 1 = 2
![\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%20%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%20%5C%20%5Ctextgreater%20%5C%20%20%5Csqrt%7B47%7D%5Csqrt%7B2%5E2%7D%20)
![\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%20%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%20%5C%20%5Ctextgreater%20%5C%20%20%5Csqrt%7B2%5E2%7D%3D2%20%5C%20%5Ctextgreater%20%5C%20%202%5Csqrt%7B47%7D%20%5C%20%5Ctextgreater%20%5C%20%20%5Cfrac%7B2%2B2%5Csqrt%7B47%7D%7D%7B2%7D)
Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.
Therefore our final solutions are
Hope this helps!
Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
Lets use the quadratic formula now
Multiply the numbers 2 * 1 = 2
Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.
Therefore our final solutions are
Hope this helps!