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2 years ago

Which of the following are among the five basic postulates of Euclidean geometry?

Mathematics

1 answer:

lana66690 [7]2 years ago

6 0

Answer:

Choices B, C, and D are correct.

Step-by-step explanation:

Euclid's Five postulates are as follows:

1. A straight line segment may be drawn from any given point to any other.

2. A straight line may be extended to any finite length.

3. A circle can be drawn with any center and radius.

4. All right angles are congruent/equal to each other.

5. Through a point not on a given line, there is exactly one line parallel to the given line.

Now looking at these postulates and the choices we are given we see that only choice B, C, and D match with above.

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Drag each tile to the table to multiply each row heading by each column heading.

Alik [6]

Answer:

Answers are below

Step-by-step explanation:

4t x 5t = 20t²

4t x -4 = -16t

5 x 5t = 25t

5 x -4 = -20

The top left box should be 20t²

The top right box should be -16t

The bottom left box should be 25t

The bottom right box should be -20

3 0

2 years ago

6 people sharing 40 lbs of gold evenly. How many do they get? Then subtract 1 2/3 from the result of the problem above & sim

r-ruslan [8.4K]

Answer: 6

Step-by-step explanation:

Given Data:

Number of people = 6

Amount of Gold to be shared = 40lbs

Subtract 2/3 from your answer.

Therefore

If six people are to share 40lbs of gold evenly

= 40lbs / 6

Divide through by 2

= 20lbs / 3

= 20lbs / 3 - 2/3

= 18/3

= 6

Each person would get approximately 6lbs of gold

7 0

1 year ago

Read 2 more answers

Mrs. Adams bought cans of pears and cans of mixed fruit.

weeeeeb [17]

Answer: 5 Cans

Step-by-step explanation:

4 0

2 years ago

Write the expression 12-2 in simplest form.

Vesnalui [34]

well, if the expression is simply 12-2, then you would do simple subtraction and your answer would be 10. that would be simplest form. you cannot take it any simpler because there is nothing else left now. your answer would be 10.

3 0

2 years ago

A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacemen

Elan Coil [88]

Answer:

When Two containers are selected

(a) Probability that the second one selected is defective given that the first one was defective = 0.00450

(b) Probability that both are defective = 0.0112461

(c) Probability that both are acceptable = 0.986

2. When Three containers are selected

(a) Probability that the third one selected is defective given that the first and second one selected were defective = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay = 0.00451

(c) Probability that all three are defective = 6.855 x $10^{-8}$ .

Step-by-step explanation:

We are given that a batch of 445 containers for frozen orange juice contains 3 defective ones i.e.

Total containers = 445

Defective ones = 3

Non - Defective ones = 442 { Acceptable ones}

Two containers are selected, at random, without replacement from the batch.

(a) Probability that the second one selected is defective given that the first one was defective is given by;

Since we had selected one defective so for selecting second the available

containers are 444 and available defective ones are 2 because once

chosen they are not replaced.

Hence, Probability that the second one selected is defective given that the first one was defective = $\frac{2}{444}$ = 0.00450

(b) Probability that both are defective = P(first being defective) +

P(Second being defective)

= $\frac{3}{445} + \frac{2}{444}$ = 0.0112461

Since, total number of acceptable containers are 442 and total containers are 445.

So, Required Probability = $\frac{442}{445}*\frac{441}{444}$ = 0.986

Three containers are selected, at random, without replacement from the batch.

(a) Probability that the third one selected is defective given that the first and second one selected were defective is given by;

Since we had selected two defective containers so now for selecting third defective one, the available total containers are 443 and available defective container is 1 .

Therefore, Probability that the third one selected is defective given that the first and second one selected were defective = $\frac{1}{443}$ = 0.002.

(b) Probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay is given by;

Since we had selected two containers so for selecting third container to be defective, the total containers available are 443 and available defective containers are 2 as one had been selected.

Hence, Required probability = $\frac{2}{443}$ = 0.00451 .

P(Second being defective) + P(Third being defective)

= $\frac{3}{445}* \frac{2}{444} * \frac{1}{443}$ = 6.855 x $10^{-8}$ .

5 0

2 years ago