Sarah must maintain a balance of at least $500 in her checking account to avoid finance charges. If her current balance is $794, write an inequality to determine how many times she can withdraw $25 for shopping without paying finance charges.

so since her balance must be at least 500, her ballance cannot reach 500 so 794-500=294 see how many 25's you can fit into 294 we know that 4 25's =100 and 294 is roughly 300 so 4 times 3=12 we minus one of the 25's because 294 is less than 300 so the innequality is

means more than 794-25x>500

Answered by apologiabiology | 2024-06-10

Let's represent the number of times Sarah can withdraw $25 as x . Sarah's current balance is $794, and she must maintain a balance of at least $500. So with each withdrawal of $25, her balance will decrease by that amount. The inequality to determine how many times she can withdraw $25 without her balance going below $500 is: 794 − 25 x ≥ 500 Now, to solve for x , we will subtract 794 from both sides of the inequality: − 25 x ≥ 500 − 794 − 25 x ≥ − 294 Now, because we are dividing by a negative number, we have to reverse the inequality sign: x ≤ − 25 − 294 ​ When we divide -294 by -25, we get a positive number: x ≤ 11.76 Since Sarah cannot withdraw a fraction of $25, we have to round down to the nearest whole number because we are looking for the maximum number of full withdrawals she can perform. So the greatest whole number less than or equal to 11.76 is 11 .
Therefore, Sarah can withdraw $25 a maximum of 11 times without her account balance dropping below $500 and incurring finance charges.

Answered by JorjaFox | 2024-06-24

Sarah can withdraw $25 from her account a maximum of 11 times without incurring finance charges, keeping her balance above $500. This is determined by solving the inequality based on her current balance of $794. The inequality found is 794 - 25x ≥ 500, leading to x ≤ 11.
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Answered by apologiabiology | 2024-12-26