Respuesta :
Answer:
[tex]\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2[/tex]
And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:
[tex] \bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0[/tex]
And we can solve for [tex]\sum_{i=1}^n w_f *X_f [/tex] and solving we got:
[tex] 3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f [/tex]
And from the previous result we got:
[tex] 3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f[/tex]
And solving we got:
[tex] \sum_{i=1}^n w_f *X_f =120 -67.2= 52.8[/tex]
And then we can find the mean with this formula:
[tex] \bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3[/tex]
So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0
Step-by-step explanation:
For this case we know that the currently mean is 2.8 and is given by:
[tex] \bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8[/tex]
Where [tex] w_i[/tex] represent the number of credits and [tex]X_i[/tex] the grade for each subject. From this case we can find the following sum:
[tex]\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2[/tex]
And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:
[tex] \bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0[/tex]
And we can solve for [tex]\sum_{i=1}^n w_f *X_f [/tex] and solving we got:
[tex] 3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f [/tex]
And from the previous result we got:
[tex] 3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f[/tex]
And solving we got:
[tex] \sum_{i=1}^n w_f *X_f =120 -67.2= 52.8[/tex]
And then we can find the mean with this formula:
[tex] \bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3[/tex]
So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0
