Answer:
c. all accounts fewer than 31 or more than 60 days past due.
Step-by-step explanation:
The Universal Set is the set of all past due accounts.
Event A = the event that the account is between 31 and 60 days past due.
Event B = the event that the account is that of a new customer.
Therefore, the complement of A will be the event of all the accounts fewer than 31 or more than 60 days past due.
The correct option is C.
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Answer:
Option B
Step-by-step explanation:
The cost of 6 turkeys at $1 per pound will be between $78 and $90. The cost of 6 turkeys at $20 each, B2G1, will be $80. We imagine that the average weight of turkeys in the 13-15 lb range will usually exceed 13 1/3 pounds, the weight at which the cost per pound is the same. That means, 6 turkeys at $1 per pound will likely cost more than $80, unless extreme effort is made to choose only the lightest turkeys.
Option B is likely to be cheaper; certainly so, if heavier turkeys are preferred.
Answer:
see explanation
Step-by-step explanation:
The translation represented by ![\left[\begin{array}{ccc}1\\4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C4%5C%5C%5Cend%7Barray%7D%5Cright%5D)
interprets as a shift of 1 unit to the right ( add 1 to x- coordinate ) and a
shift of 4 units down ( subtract 4 from the y- coordinate ), then
(1, 4 ) → (1 + 1, 4 - 4 ) → (2, 0 )
(4, 4 ) → (4 + 1, 4 - 4 ) → (5, 0 )
(6, 2 ) → (6 + 1, 2 - 4 ) → (7, - 2 )
(1, 2 ) → (1 + 1, 2 - 4 ) → (2, - 2 )
Answer:
The factored expression is 2(x² + 5)(x + 3).
Step-by-step explanation:
Hey there!
We can use a factoring technique referred to as "grouping" to solve this problem.
Grouping is used for polynomials with four terms as a quick and easy factoring method to remove the GCF and get down to the initial terms that create the expression/function.
Grouping works in the following matter:
- Given equation: ax³ + bx² + cx + d
- Group a & b, c & d: (ax³ + bx²) + (cx + d)
- Pull GCFs and factors
Let's apply these steps to the given equation.
- Given equation: 2x³ + 6x² + 10x + 30
- Group a & b, c & d: (2x³ + 6x²) + (10x + 30)
- Pull GCFs and factors: 2x²(x + 3) + 10(x + 3)
As you'll see, we have a common term with both sides of the expression. This term, (x + 3), is a valuable asset to the factoring process. This is one of the factors for our expression.
Now, we use our GCFs to create another factor.
- List GCFs: 2x², 10
- Create a term: (2x² + 10)
Finally, we'll need to simplify this one by taking another GCF, 2.
- Pull GCF: 2(x² + 5)
Now that we have this term, we need to understand that this <em>could</em> also be factored further using imaginary numbers, but it is also acceptable to leave it in this form.
Therefore, we have our final factors: 2(x² + 5) and (x + 3).
However, when we factor, we place all of our terms together. This leaves us with the final answer: 2(x² + 5)(x + 3).
