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chubhunter [2.5K]
3 years ago
7

Which statement is true about this argument ?

Mathematics
2 answers:
MrRa [10]3 years ago
5 0
D the argument is not valid because the conclusion does not follow from the premises
nignag [31]3 years ago
5 0
It's good to review the laws of syllogism and detachment.

Law of detachment:
Statement 1: If p then q.
Statement 2: p is true.
By the law of detachment, you can conclude "q is true."

Law of syllogism:
Statement 1: If p then q.
Statement 2: If q then r.
By the law of syllogism, you can conclude "If p then r."

Now look at which of the two cases above you have.
Statement 1: If a quadrilateral is a square, then the quadrilateral is a rectangle.
This is "if p then q."

Statement 2:
<span>If a quadrilateral is a rectangle then the quadrilateral is a parallelogram.
This is "if q then r."

You have
If p then q.
If q then r.

This is what you need for the law of syllogism.

That means you can conclude "if p then r", which in this specific case is
"</span><span>If a quadrilateral is a square, then the quadrilateral is a parallelogram."
</span>
The answer is that it is a valid argument by the law of syllogism.
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What is the equivalent fraction of 4/5​
Aleks04 [339]

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16/20

Step-by-step explanation:

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What is similar about a line segment and a line? What is different
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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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