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3 years ago

Three student attempt to define what a line segment it?

SAT

1 answer:

ser-zykov [4K]3 years ago

4 0

Answer:

First definition:

Duru's definition: A line segment is a line connecting two endpoints

⇅

Teacher's comment: What you've written is true, but it isn't a correct definition. A ray would also fit your defenition

Second definition:

Oliver's definition: A line segment is when three points are all on the same line.

⇅

Teacher's comment: Were you thinking of the team collinear?

Third definition:

Carlos' definition: A line segment is<u> part of a line,</u> not the whole thing.

⇅

Teacher's comment: I think you have the right idea, but you need to be more careful with your language.

Explanation:

A line segment is <u>part of a line</u> that connects points. It limited consequently, it has length.

Duru's definition: A line segment is a line connecting two endpoints

It is missing that it is just a part of the line. A ray also connects two points but extends infinetly in one direction. Thus, Duru's definition is not so specific.

That justifies teacher's comment:

Teacher's comment: What you've written is true, but it isn't a correct definition. A ray would also fit your defenition

Second definition:

Oliver's definition: A line segment is when three points are all on the same line.

The constraint of a line segment to three points all on the same line corresponds to collinear points. Two points always form a line,, but three points lie on a line only in they are collinear.

That justifies, theacher's comment:

Teacher's comment: Were you thinking of the team collinear?

Third definition:

Carlos' definition: A line segment is part of a line, not the whole thing.

Mathematicall language is formal. You must use precise words. An expression as 'whole thing" is not formal.

That justifies teacher's comment:

Teacher's comment: I think you have the right idea, but you need to be more careful with your language.

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2 years ago

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3 years ago

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3 years ago

Consider a set of cards that has four cards labeled 1, 2, 3, and 4. Suppose you pick two cards, without replacement, to obtain t

Phantasy [73]

Answer:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

Explanation:

Given

$Cards = \{1,2,3,4\}$

Required

The sampling distribution

The possible selection of 2 cards without replacement is as follows:

$S = \{(1,2) (1,3) (1,4) (2,1) (2,3) (2,4) (3,1) (3,2) (3,4) (4,1) (4,2) (4,3)\}$

Calculate the mean

$\begin{array}{cccccccccccc}{Selection} & {(1,2)} & {(1,3)} & {(1,4)} & {(2,1)} & {(2,3)} & {(2,4)}& {(3,1)} & {(3,2)} & {(3,4)} & {(4,1)} & {(4,2)} & {(4,3)} \ \\ {Mean} & {1.5} & {2} & {2.5} & {1.5} & {2.5} & {3} & {2} & {2.5} & {3.5} & {2.5} & {3} & {3.5}\ \end{array}$

List out the mean and the respective frequency

$1.5 \to 2$

$2 \to 2$

$2.5 \to 4$

$3 \to 2$

$3.5\to 2$

$Total \to 12$

Calculate the probability of each mean

$P(1.5) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2.5) \to \frac{4}{12} \to \frac{1}{3}\\$

$P(3) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(3.5) \to \frac{2}{12} \to \frac{1}{6}$

So, the table of sampling distribution is:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

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3 years ago