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2 years ago

A, B, and C are collinear, and B is between A and C. The ratio of AB to BC is 3 : 1.

Mathematics

1 answer:

S_A_V [24]2 years ago

4 0

Coordinates of point C: (1,-1)

Step-by-step explanation:

In this problem, A, B and C are collinear, and B is between A and C.

The ratio AB : BC is 3 : 1.

This means that we can write the following two equations:

$x_B-x_A = 3(x_C-x_B)\\y_B-y_A=3(y_X-y_B)$

where:

$(x_A,y_A)=(-7,3)$ are the coordinates of point A

$(x_B,y_B)=(-1,0)$ are the coordinates of point B

$(x_C,y_C)$ are the coordinates of point C

Solving the equation for $x_C$ ,

$x_C = x_B + \frac{x_B-x_A}{3}=-1+\frac{-1-(-7)}{3}=1$

Solving the equation for $y_C$ ,

$y_C=y_B + \frac{y_B-y_A}{3}=0+\frac{0-3}{3}=-1$

So, the coordinates of point C are

$C(-1,1)$

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Lannister checks out 6 set of books from the library each set has the same number of books she reads 7 of the books and she has

fgiga [73]

Answer:

Step-by-step explanation:

Lannister checks out 6 sets of book

Each set had the same number of books

She reads 7 books and has 11 more to go

Therefore the number of books in each set can be calculated as follows

= 11+7

= 18

18/6

= 3

Hence there are 3 books in each set

6 0

3 years ago

you currently have 24 credit hours and a 2.8 gpa you need a 3.0 gpa to get into the college. if you are taking a 16 credit hours

Juliette [100K]

Answer:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

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:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

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:

$\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8$

Where $w_i$ represent the number of credits and $X_i$ the grade for each subject. From this case we can find the following sum:

$\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2$

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:

$\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0$

And we can solve for $\sum_{i=1}^n w_f *X_f$ and solving we got:

$3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f$

And from the previous result we got:

$3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f$

And solving we got:

$\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8$

And then we can find the mean with this formula:

$\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3$

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

6 0

2 years ago

a class with 30 students had an average score of 80 on a test. A class with 20 students had an average score of 90 on the same t

snow_tiger [21]

The average score of all the students in both classes is 84.

<h3>How to calculate the average?</h3>

The class with 30 students had an average score of 80 on a test. The total score will be:

= 30 × 80

= 2400

A class with 20 students had an average score of 90 on the same test. The total score will be:

= 20 × 90

= 1800

Total scores = 2400 + 1800 = 4200

Number of students = 20 + 30 = 50

The average score will be:

= Total score / Total students

= 4200 / 50

= 84

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Vinvika [58]

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2 years ago

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gayaneshka [121]

Answer:

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