Respuesta :
Answer:
[tex]100 + 20n \leq 230[/tex]
[tex]n \leq 6[/tex]
Step-by-step explanation:
Given
[tex]Balance = \$230.00[/tex]
[tex]Minimum = \$100[/tex]
[tex]Withdrawal = \$20\ each[/tex]
Required
Represent using an inequality and solve
Let the number of dollars be represented by n
If 1 withdrawal is $20 bill
n withdrawals would be $20n bills
The total withdrawal and the minimum balance must be less than or equal to the available balance.
So, this gives
[tex]Minimum + Total\ Withdrawal \leq Available\ Balance[/tex]
Substitute values for each of these
[tex]100 + 20n \leq 230[/tex]
Solving for n
[tex]20n \leq 230 - 100[/tex]
[tex]20n \leq 130[/tex]
Divide through by 20
[tex]n \leq 6.5[/tex]
But n can not be decimal because it represents number of dollar bills , which must always be an integer.
So, we round down the value of n
[tex]n \leq 6[/tex]
Here, we are required to identify and solve the inequality that represents how many $20 bills, n that can be withdrawn from the account without going below the minimum balance where current balance = $230.00 and minimum balance = $100.
The required inequality is;
$230 - $20n ≥ $100
And the number of $20 bills that can be withdrawn without going below the minimum balance is 6.5 (i.e 6, since the $20 come out whole).
From the statements above,
- Current balance = $230
- Withdrawal is in $20bills denomination occuring, n times.
- The minimum balance = $100.
We can form an inequality about the situation as follows;
$230 - $20n ≥ $100
To solve the inequality, we have;
-$20n ≥ $100 -$230
-$20n ≥ -$130
Therefore, n ≤ 130/20
Therefore, n ≤6.5
Ultimately, the number of $20 bills that can be withdrawn without going below the minimum balance is 6.5 (i.e 6, since the $20 come out whole).
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