All categories

2 years ago

Geometry (19) Study Guide My answer: x = 11 so 2(11) - 6 = 16, FG = 16

Mathematics

1 answer:

jok3333 [9.3K]2 years ago

3 0

I think you meant to say n = 11. If so, then you are correct and it leads to the proper length for FG, which is 16 units. Nice work.

You might be interested in

.The endpoints of AB are A(1,4) and B(6,-1).

Brums [2.3K]

Answer:

5√2

Step-by-step explanation:

Finding distance :-

d = √{ ( 1 -6)² + (4+1)²}
d = √{ -5² + 5² }
d = √{ 25 + 25}
d = √50
d = 5√2

4 0

2 years ago

Solve this problem -282 - (+1,017)=2

77julia77 [94]

The answer is not true.

6 0

2 years ago

What multiply what equals 26 using numbers between 2 and 9?

nignag [31]

There are only 1*26 and 2*13 so there is no answer unless you allow decimals.

8 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

Let

Citrus2011 [14]

Answer:

hi or high

Step-by-step explanation:

so you right the world h and i and cobine them to make hi or high same thing

3 0

2 years ago