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11Alexandr11 [23.1K]

1 year ago

9

You are shopping for disposable camera, to hand out at a party. the daytime camera costs $2.75 and the flash camera costs $4.25.

you must buy exactly 20 cameras and you want to spend between $65 and $75, inclusive. write and solve a compound inequality for this situation. list all the solutions that involve whole numbers of cameras.

1 answer:

Svetach [21]1 year ago

6 0

The solutions that involve whole numbers of cameras are (7, 13), (8, 12), (9, 11), (10, 10), (11, 9), (12, 8) and (13, 7)

<h3>Write and solve a compound inequality for this solution.</h3>

The given parameters are:

Daytime cameras = $2.75
Flash camera = $4.25
Number of cameras = 20 cameras
Amount spent = Between $65 and $75

Represent the daytime camera with x and the flash camera with y.

So, we have:

x + y = 20

65 <=2.75x + 4.25y <= 75

Make y the subject in x + y = 20

y = 20 - x

Substitute y = 20 - x in 65 <=2.75x + 4.25y <= 75

65 <=2.75x + 4.25(20 - x) <= 75

Expand

65 <=2.75x + 85 - 4.25x <= 75

This gives

65 <=-1.5x+ 85 <= 75

Subtract 85 from the inequality

-20 <=-1.5x <= -10

Split the inequality

-1.5x >= -20 and -1.5x <= -10

Divide by -1.5

x <= 13.33 and x >= 6.67

Combine the inequality

6.67 <= x <= 13.33

<h3>List all of the solutions that involve whole numbers of cameras.</h3>

In (a), we have:

6.67 <= x <= 13.33

This means that:

x = 7, 8, 9, 10, 11, 12, 13

Recall that:

y = 20 - x

So, we have:

y = 13, 12, 11, 10, 9, 8, 7

So, the solutions that involve whole numbers of cameras are (7, 13), (8, 12), (9, 11), (10, 10), (11, 9), (12, 8) and (13, 7)

Read more about inequality at:

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