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Rama09 [41]
2 years ago
11

There were 24 children hiding. Sonya found 3/4 of them. How many children did she fine?

Mathematics
2 answers:
Vera_Pavlovna [14]2 years ago
6 0

Answer:

Then Sonya only found 18 children

Step-by-step explanation:

Because 24/4=6

And 6*3=18 Children

So the answer is 18 Children

<em>Hope I Helped</em>

Rzqust [24]2 years ago
4 0

Answer:

She found 18 of them

Step-by-step explanation:

3/4 of 24 is 18

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Let C(x) be the statement "x has a cat," let D(x) be the statement "x has a dog," and let F(x) be the statement "x has a ferret.
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Answer:

\mathbf{a)} \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\\\mathbf{b)} \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee \; F(x)\\\mathbf{c)} \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)\\\mathbf{d)} \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)\\\mathbf{e)} \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

Step-by-step explanation:

Let X be a set of all students in your class. The set X is the domain. Denote

                                        C(x) -  ' \text{$x $ has a cat}'\\D(x) -  ' \text{$x$ has a dog}'\\F(x) -  ' \text{$x$ has a ferret}'

\mathbf{a)}

Consider the statement '<em>A student in your class has a cat, a dog, and a ferret</em>'. This means that \exists x \in X so that all three statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                         \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)

\mathbf{b)}

Consider the statement '<em>All students in your class have a cat, a dog, or a ferret.' </em>This means that \forall x \in X at least one of the statements C(x), D(x) and F(x) is true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                        \left( \forall x \in X\right) \; C(x) \; \vee \; D(x) \; \vee F(x)

\mathbf{c)}

Consider the statement '<em>Some student in your class has a cat and a ferret, but not a dog.' </em>This means that \exists x \in X so that the statements C(x), F(x) are true and the negation of the statement D(x) . We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as follows

                      \left( \exists x \in X\right) \; C(x) \; \wedge \; F(x) \; \wedge \left(\neg \; D(x) \right)

\mathbf{d)}

Consider the statement '<em>No student in your class has a cat, a dog, and a ferret..' </em>This means that \forall x \in X none of  the statements C(x), D(x) and F(x) are true. We can express that in terms of C(x), D(x) and F(x) using quantifiers, and logical connectives as a negation of the statement in the part a), as follows

\neg \left( \left( \exists x \in X\right) \; C(x) \; \wedge \; D(x) \; \wedge \; F(x)\right) \iff \left( \forall x \in X\right) \; \neg C(x) \; \vee \; \neg D(x) \; \vee \; \neg F(x)

\mathbf{e)}

Consider the statement '<em> For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.' </em>

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           \left((\exists x\in X)C(x) \right) \wedge  \left((\exists x\in X) D(x) \right) \wedge \left((\exists x\in X) F(x) \right)

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3 years ago
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