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Juli2301 [7.4K]
3 years ago
14

The following argument is an instance of one of the five inference forms MP, MT, HS, DS, Conj. Identify the form.

Mathematics
1 answer:
wel3 years ago
5 0

Answer:

e. HS

Step-by-step explanation:

The argument:

[(P ≡ T) • (H • N)] ⊃ (T ⊃ ~S)

(T ⊃ ~S) ⊃ [(H ∨ E) ∨ R]

[(P ≡ T) • (H • N)] ⊃ [(H ∨ E) ∨ R]

is an instance of one of hypothetical syllogism (HS).

Hypothetical syllogism contains conditional statements for its premises.

Let

p = [(P ≡ T) • (H • N)]

q = (T ⊃ ~S)

r = [(H ∨ E) ∨ R]

The this can be interpreted as:

p ⊃ q

q ⊃ r

p ⊃ r

This interprets that:

If p then q

but if q then r

therefore if p then r

Thus, in logic HS is a valid argument form:

p → q

q → r

∴ p → r

Note that ⊃ symbol is used to symbolize implication relationships. This is used in conditional statements which are represented in the if...then... form.  For example p ⊃ q means: if p then q. So the type of Hypothetical syllogism used in this is conditional syllogism.

There are three parts of syllogism:

major premise

minor premise

conclusion

An example is:

If ABC is hardworking, then ABC will go to a good college.  

Major premise: ABC is hardworking.

Minor premise: Because ABC is hardworking , ABC will score well.

Conclusion: ABC will go to a good college.

Example of Hypothetical syllogism:

If AB is a CD, then EF is a GH

if WX is a YZ, then AB is a CD

therefore if WX is a YZ, then EF is a GH

This can be understood with the help of an example:

If you study the topic, then you will understand the topic.  

If you understand the topic, then you will pass the quiz.

Therefore, if you study the topic, then you will pass the quiz.

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Answer:

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3 years ago
The ratio of the number of peas Julia had to the number of peas Vlada had was 3:2. After Julia gave Vlada 15 peas, she still had
inn [45]
J:V = 3:2

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3 years ago
5. 3x + 5y + 5z =1<br> x - 2y = 5<br> 2x + 4y = 11
lilavasa [31]

Answer:

see explanation

Step-by-step explanation:

Given the 3 equations

3x + 5y + 5z = 1 → (1)

x - 2y = 5 → (2)

2x + 4y = 11 → (3)

Use (2) and (3) to solve for x and y

Multiply (2) by 2

2x - 4y = 10 → (4)

Add (3) and (4) term by term

4x = 21 ( divide both sides by 4 )

x = \frac{21}{4\\}

Substitute this value of x into (3)

2 × \frac{21}{4\\} + 4y = 11

\frac{21}{2\\} + 4y = 11 ( subtract \frac{21}{2\\} from both sides )

4y = \frac{1}{2} ( divide both sides by 4 )

y = \frac{1}{8\\}

Substitute the values of x and y into (1) and solve for z

3 × \frac{21}{4\\} + 5 × \frac{1}{8\\} + 5z = 1

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\frac{131}{8} + 5z = 1 ( subtract \frac{131}{8} from both sides )

5z = - \frac{123}{8} ( divide both sides by 5 )

z = - \frac{123}{40}

Solution is

x = \frac{21}{4\\}, y = \frac{1}{8\\}, z = - \frac{123}{40}

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Solve this equation by using the substitution method. Make sure you show ALL your work.
djverab [1.8K]

Step-by-step explanation:

substitute for y

x+3+4 = -6x

x+ 7 = -6x

7 = -7x

-1 = x

y = -1+3

y = 2

4 0
3 years ago
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