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**Hint:**We know that if we have two predictions p and q, then the implication $p\Rightarrow q$ is true when either p is false, or q is true, or both. Also, we must know that a conjunction is true only when both the predictions are true. Using these two concepts, we can show that $p\wedge q\Rightarrow r$ is true when p is true and q is false.

**Complete step by step answer:**

We know that if we have two predictions p and q, then $p\Rightarrow q$ is true when either p is false, or q is true, or both.

We also know very well that the conjunction $p\wedge q$ is true only if both p and q are true, else the conjunction is false. We can also see this from the truth table shown below,

p | q | $p\wedge q$ |

True | True | True |

True | False | False |

False | True | False |

False | False | False |

In this question, we are given that the predicted p is true and the predicted q is false.

From the truth table, we can see that when p is true and q is false, then the conjunction $p\wedge q$ is false.

And from the above discussion on implication, we can say that the implication $p\wedge q\Rightarrow r$ is true when either $p\wedge q$ is false, or r is true, or both.

We have no information about r. Now, we have proved above that the conjunction $p\wedge q$ is false.

So, we can easily say that the implication $p\wedge q\Rightarrow r$ is true.

Hence, when p is true and q is false, the implication $p\wedge q\Rightarrow r$ is true.

**Note:**We must remember that an implication is also called a conditional. We should also know that the conjunction, represented by the symbol $\wedge $, is also called AND logic. We can also solve this problem using negation and disjunction.

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