Rob collects data about how many customers enter and leave a store every hour. He records a positive number for customers enteri
ng the store and a negative number for customers leaving the store each hour. a. During which hour did more customers leave the most?
b. There were 75 customers in the store at 1:00. The store must be emptied of customers when it closes at 5:00. How many customers must leave the store between 4:00 and 5;00?
Entering Leaving
1:00 to 2:00 30 -12
2:00 to 3:00 14 -8
3:00 to 4:00 18 -30
To answer the question, we utilize the given data and solve for the number of customers left in the store after every 1 hour interval.
Initially, there were 75 customers.
1:00 to 2:00 : 75 + 30 - 12 = 93 2:00 to 3:00 : 93 + 14 - 8 = 99 3:00 to 4:00 : 99 + 18 - 30 = 87 4:00 to 5:00 : If the store does not accept any more customers except those who are already inside then, all 87 must leave between 4:00 to 5:00 in order to empty out the store by 5:00
Distribute 6 to 5/8u and 1 and you get 15/4u + 6. Distribute -6 to -7/4u and -5 and you get 21/2u and 30. The equation will be 15/4u + 6 + 21/2u + 30 and then combine like terms to get 57/4u + 36.
Because you cannot add the two numbers you must find a common denominator for 16 and 20. This answer is 4. The greatest number of group members possible is 4.
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