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geniusboy [140]
4 years ago
5

The optimal depth (d) for catching a certain type of fish satisfies the inequality 28|d – 350| – 7000 < 0. Which solution rep

resents the range of depth that offers the best catch?
A 250 < d < 300
B 100 < d < 600
C 300 < d < 350
D 10 < d < 400
Mathematics
1 answer:
Bess [88]4 years ago
3 0
<span>28|d – 350| – 7000 < 0

add 7000 to each side

</span>28|d – 350|  < 7000
<span>
divide both sides by 28

</span>|d – 350|  < 250

remove absolute value term on left so the right side becomes +/-
d-350 < +/- 250

now add 250 to 350 and subtract 250 from 350

250 +350 = 600
350-250 = 100

answer is: <span>B 100 < d < 600</span><span>



</span>
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Answer:

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Step-by-step explanation:

2x14= 28

28-9=19

19-17=2

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Step-by-step explanation:

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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

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3 years ago
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seraphim [82]

Ooo! I love these! Ok so your IQR is C or D. I might have my calculations wrong if so. Sorry!

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