Answer:
[tex]0.10d + 0.25q > 143.88[/tex]
Step-by-step explanation:
Let the total dimes collected be d.
[tex]1\ dime = \$0.10[/tex]
[tex]d\ dimes = \$0.10d[/tex]
Let the total quarters collected be q.
[tex]1\ quarter = \$0.25[/tex]
[tex]q\ quarters = \$0.25q[/tex]
[tex]Last\ Year = \$143.88[/tex]
Required
Represent as an inequality
The total collection, this year, can be represented as:
[tex]Total = 0.10d + 0.25q[/tex]
From the question, we understand that:
This year's collections is expected to be greater than last year's
This can be represented as:
[tex]Total > Last\ Year[/tex]
Substitute in the right values:
[tex]0.10d + 0.25q > 143.88[/tex]