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PolarNik [594]
3 years ago
15

Mary has twice as many dogs as David has cats and josh has enough animal collars to give a collar to one third of the dogs or to

give a collar to all the cats except one. How many collars does josh have?
Mathematics
1 answer:
Ann [662]3 years ago
3 0
D = 2C
J = 1/3D
J = C - 1

J = 1/3D
J = 1/3(2C)
J = 2/3C

2/3C = C - 1
2/3C - 3/3C = -1
-1/3C = -1
C = -1 * -3
C = 3...there are 3 cats

J = C - 1
J = 3 - 1
J = 2.....there are 2 collars





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 Let X be a discrete binomial random variable.
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Which expression can be used to determine the length of segment AB? On a coordinate plane, triangle A B C has points (4, 3), (ne
JulijaS [17]

Answer:

StartRoot 2 squared + 6 squared EndRoot

Step-by-step explanation:

we have

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we know that

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substitute the given values

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3 years ago
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Ksivusya [100]

Answer:

Yes, an arrow can be drawn from 10.3 so the relation is a function.

Step-by-step explanation:

Assuming the diagram on the left is the domain(the inputs) and the diagram on the right is the range(the outputs), yes, an arrow can be drawn from 10.3 and the relation will be a function.

The only time something isn't a function is if two different outputs had the same input. However, it's okay for two different inputs to have the same output.

In this problem, 10.3 is an input. If you drew an arrow from 10.3 to one of the values on the right, 10.3 would end up sharing an output with another input. This is allowed, and the relation would be classified as a function.

However, if you drew multiple arrows from 10.3 to different values on the right, then the relation would no longer be a function because 10.3, a single input, would have multiple outputs.

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Consider the following statement. ∀ integer d, if 6 d is an integer then d = 3. Which of the following is a negation for the sta
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Answer:

The correct option is the first:

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Step-by-step explanation:

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Let "If 6d is an integer" be p and "then d = 3" be q.

The statement is a conditional statement. Therefore, we have

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The negation of p ⇒q, ¬(p ⇒ q) ≡ p ∧ ¬q

I have attached an image proving the above.

Using this: ¬(p ⇒ q) ≡ p ∧ ¬q

Then, we leave p as it is and negate q.

P is "if 6d is an integer"

q is "then d = 3," hence ¬q is "then d ≠ 3."

Therefore, the negation of "If 6d is an integer, then d = 3" is "If 6d is an integer, then d ≠ 3."

The correct option is the first:

∃ an integer d such that 6d is an integer and d ≠ 3

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