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Dafna1 [17]
3 years ago
12

There are 12 green,22 white, 12 purple and 4 blue gum balls in the gum all machine what is the probability of getting white gum

all write the probability as a percent
Mathematics
1 answer:
NemiM [27]3 years ago
4 0

Answer: If you add all of these numbers of gumballs up you get 50. Since there is 50 gumballs,and 100 percent of that amount has to be filled,then you must make each gumball worth 2 percent. (12x2)+(22x2)+(12x2)+(4x2)=100

Green percent of getting=24 percent

Purple percent of getting=24 percent

Blue percent of getting=8 percent

White percent of getting=44 percent

The percent of getting a White gum is 44 percent.

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\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)}

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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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