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:
brainly.com/question/25275758
#SPJ1
