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artcher [175]
3 years ago
13

Dorian buys 2 pounds of almonds for $21.98 and 3 pounds of dried apricots for $26.25. Which is less expensive per pound? How muc

h less expensive?
Mathematics
1 answer:
ozzi3 years ago
4 0

Answer:

The apricots are less expensive per pound.

Step-by-step explanation:

Cost of 1 pound of almonds = 21.98/2

Cost of 2 pounds of almonds = $21.98

So then you have to do 21.98 - 21.98/2. Which equals $10.99

Cost of 1 pound of dried apricots = 26.25/3

Cost of 3 pounds of dried apricots = $26.25

So then you have to do $26.25 - 26.25/3 which = $17.5 then divide that by 2 which equals $8.75.

To find out how much less expensive you have to find the difference, so to do that you have to do $10.99 - $8.75 = $2.24 per pound less expensive.

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I really need help with this ?
Digiron [165]
The answer is g. 47.1

you use the formula
c= π d
then plug is 15 for d and you should get the answer
8 0
3 years ago
D= 7/2 cm
kvv77 [185]

Answer:

1) B

2) C

3) B

Step-by-step explanation:

Radius = D/2 = 7/2 × 2 = 7/4

Circumference = pi × D

= 7/2 pi

7/2 × 3.14 = 10.99

Approximately 11

3 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
3 years ago
I need help solving this
pentagon [3]
Y=mx+b where m=slope=(dy/dx) and b=y-intercept (value of y when x=0)

You have two points...(1,-2),(3,2) so

m=(y2-y1)/(x2-x1)=(2--2)/(3-1)

m=4/2=2 now that you have the slope...

y=2x+b, you can use either point to solve for the y-intercept, I'll use (3,2)

2=2(3)+b

2=6+b

b=-4 so the line is:

y=2x-4
5 0
3 years ago
Which choice is equivalent to expresision below?
DIA [1.3K]

We can simplfiy to be -\sqrt{100}

\sqrt{-1} * \sqrt{100}.

We know that \sqrt{100} is 10 and -1 is i, so the answer is 10i, or D.


5 0
3 years ago
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