Premises: All good students are good readers. Some math students are good students. Conclusion: Some math students are good read
ers. Use the law of detachment or the law of contraposition to form a valid conclusion from each set of premises in exer cises
1 answer:
Answer:
Therefore, the conclusion is valid.
The required diagram is shown below:
Step-by-step explanation:
Consider the provided statement.
Premises: All good students are good readers. Some math students are good students.
Conclusion: Some math students are good readers.
It is given that All good students are good readers, that means all good students are the subset of good readers.
Now, it is given that some math students are good students, that means there exist some math student who are good students as well as good reader.
Therefore, the conclusion is valid.
The required diagram is shown below:
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Answer:
The answer is 9.725
Step-by-step explanation:
We have been given the mixed fraction
First we will concert this to improper fraction.
=
Now we will divide this using long division or calculator and get the answer 9.725.
So, you had done everything right so far (other than squaring the 2), but that was only half of the question.
to find the least common multiple, you need to first figure out what the prime factors have in common.
each have two twos. both have one 5, so we know our answer will look something like
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.
one has one 3 and one has two, so we need two threes. now our equation is
what's the only number we have to deal with? 7...
how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes
to find the least common multiple, you need to first figure out what the prime factors have in common.
each have two twos. both have one 5, so we know our answer will look something like
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.
one has one 3 and one has two, so we need two threes. now our equation is
what's the only number we have to deal with? 7...
how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes