The number of shoppers in a store is modeled by y=-0.5|t-288|+144 where t is the time (in minutes) since the store opened at 10: 00ama. for what values are there 100 shoppers in the b. at what times are there 100 shoppers in the c. what is the greatest number of shoppers in the d. at what time does the greatest number of shoppers

... y = -0.5|t -288| +144

... t is minutes after 10 am

... a) t such that y = 100

... b) clock time corresponding to those values of t

... c) t such that y is a maximum

... d) clock time corresponding to that maximum

a) y = 100

... -0.5|t -288| +144 = 100 . . . . . substitute for y

... -0.5|t -288| = -44 . . . . . . . . . . subtract 144

... |t - 288| = 88 . . . . . . . . . . . . . multiply by -2

... -88 = t -288 . . . . . . . one meaning of the above equation

... t = 200 . . . . . . . . . . . add 288

... t -288 = 88 . . . . . . . . the other meaning of the above equation

... t = 376 . . . . . . . . . . . add 288

There are 100 shoppers in the store at t=200 and t=376

b) On a 24-hour clock, the clock time (h) corresponding to a value of t is ...

... h(t) = 10 + t/60

... h(200) = 10 + 200/60 = 13 1/3 . . . . corresponds to 1:20 pm

... h(376) = 10 + 376/60 = 16 4/15 . . . . corresponds to 4:16 pm

c) The value of y is a maximum where the content of the absolute value is 0. Thus the maximum number of shoppers is 144.

d) The maximum number of shoppers occurs when t=288, so at clock time ...

... h(288) = 10 + 288/60 = 14 4/5 . . . . corresponds to 2:48 pm

An hour is 60 minutes, so 1/3 hour is ...

... 1/3×60 = 20 minutes; and ...

... 4/15×60 = 16 minutes;

... 4/5×60 = 48 minutes.

it would be 6xy and -16xy.

answer: it's upside down and i can't read it

step-by-step explanation: