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3 years ago

I WAS ONLY ABLE TO SELECT 40 PTS THIS TIME, PLS ANSWER THIS ONE IT HAS THE PICTURE WITH IT.

Mathematics

2 answers:

lara31 [8.8K]3 years ago

6 0

∠1 and 105° are supplementary angles. (supplementary = 180°)

180 - 105 = 75°

∠1 = 75°

∠1 and ∠2 are corresponding angles, and so there measurements will be the same

∠1 = ∠2
∠1 = 75°
∠2 = 75°

105° and ∠3 are same side, and are corresponding angles. Therefore, ∠3 = 105°

m∠1=75°

180° (the measurement of a straight angle).

Subtract the 105° (given) from the straight angle

180 - 105 = 75

m∠1=75°

hope this helps

BaLLatris [955]3 years ago

4 0

Okay so the first one is right.
The second one is wrong.
the third one is right.
The forth one is wrong.

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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

3 years ago