Can you solve the inequality for: y + 13 < 4
1 answer:
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The first digit is one of {1, 3, 5, 7, 9}, and the last digit is one of {1, 2, 3, 4, 5}.
If the first digit is one of {1, 3, 5} (3 choices), then the last digit is one of {1, 2, 3, 4, 5} minus whatever is picked for the first digit (4 choices, and the remaining eight digits can be arranged in 8! ways, so there are 12*8! possible permutations.
If the first digit is one of {7, 9} (2 choices, then the last digit is one of {1, 2, 3, 4, 5} (5 choices), and again there are 8! ways of arranging the remaining digits, so there are 10*8! possible permutations.
Then the total number of permutations that fit the criteria is 12*8! + 10*8! = 22*8!. There are 10! total permutations that can be made overall, so the probability of randomly picking one we want is
We have two products: a two-slice toaster and a four-slice toaster.
Lets call T the number of two-slice toasters produced and F the number of four-slice toasters produced.
If we have 300 hours per week of labor, the sum of the labor required for T and F has to be lower or equal to that value.
The labor hours that take to make the two-slice toasters can be written as 6*T, as it is the product of the hours per unit (6 hours/unit) by the number of units (T units).
The same can be done for F, where the labor required can be expressed as 10*T.
If we add these terms, and make them be less or equal than 300, we get:
The production capacity is 40 units, so the sum of T and F has to be less or equal than 40.
We can express this as:
Answer:
The system of inequalities for this problem becomes:
6T+10F<=300
T+F<=40