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A. Jenny is looking for the best option for daycare for her son. She lives in a small town, so her options are limited to two daycare centers. Option A is a home-based facility which charges \$9.00 per hour that the child is at the home. Option B is a center-based facility which charges \$150 for the first 20 hours and then charges \$10 per hour thereafter. Due to a new promotion, Jenny will be required to work more hours away from home. Which facility will cost less for Jenny to have her son in daycare for a work week (40 hours)?

B. I. h = how many hours of child care is needed = 40 f = the flat hourly rate charge t = total amount due for the week

Option A: f * h = t a. 9 * 40 = \$360 Option B: 150 + f(h-20) = t, for h >=20 b. 150 + 10(40-20) = \$350

150 = t, for h < 20 c. 150 = total

II. I set the hours to "h", the hourly fee to "f" and the total to "t". a. Option A: This option only has an hourly fee for each hour. The fee is \$9 and the amount of hours is 40 so I multiplied the hours by the fee. b. Option B: This option has a flat rate of \$150 for the first 20 hours and then an hourly rate of \$10 thereafter. For the first equation that refers to over 20 hours, I subtracted 20 hours from the total time, multiplied it by \$10, and then added it to \$150. For the second equation that refers to any amount less than 20 hours, the total is always going to be \$150.

III. 9h = 150 9 9 h = 16.67

Set the equations equal to each other for under 20 hours. Divide by 9 to get h = 16.67 – the daycare centers are equal in price at 16.67 hours. This solution point is at (16.67, 150).

9h = 150+10(h-20) 9h = 150+10h-200 -10h - 10h -1h = 150 - 200 -1h…...

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