X f(x) -2 -1/200 -1 -1/20 0 -1/2 1 -5 2 -50
1 answer:
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Answer:
The minimum acceptable weight of the flour bag is 39.5 kg,
and its maximum acceptable weight is 40.5 kg
Step-by-step explanation:
Notice that the selling weight of the bag to be sold (x) needs to differ from 40 kg at most 0.5 kg. Therefore, we can write the following inequality:
The inequality can be solve once we remove the absolute value symbol, and solve for the unknown "x". Recall that in order to solve it, we need to consider the two possible cases:
a) that the expression within the absolute value symbol is larger than or equal to zero, and b) that the expression is less than zero.
a) If then its absolute value is equal to itself (x-40), and when we remove the absolute value symbols, we get:
which means that the weigh "x" of the flour bag should be smaller or equal than 40.5 kg
b) If , then the absolute value of this is its opposite : -x+40, and the inequality becomes:
which means that the weight of the flour bag must be larger than or equal to 39.5 kg
So, the minimum acceptable weight of the flour bag is 39.5 kg, and the maximum acceptable weight is 40.5 kg