Identify and solve the inequality that represents how many $20 bills, n, you can withdraw from the account without going below the minimum balance.
your current balance is $230.00
minimum balance is $100

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Here, we are required to identify and solve the inequality that represents how many $20 bills, n that can be withdrawn from the account without going below the minimum balance where current balance = $230.00 and minimum balance = $100.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And the number of $20 bills that can be withdrawn without going below the minimum balance is 6.5 (i.e 6, since the $20 come out whole).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Ultimately, the number of $20 bills that can be withdrawn without going below the minimum balance is 6.5 (i.e 6, since the $20 come out whole).

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