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Crank
3 years ago
13

Which biconditional statement is true?

Mathematics
2 answers:
tatyana61 [14]3 years ago
6 0

Answer: A shape is a rectangle if and only if the shape has exactly four sides and four right angles.

Step-by-step explanation:

  • If both a conditional statement and  its converse statement is true then we write a combine form of both the statements known as a bi-conditional statement. It is written in the if and only if form.

From the given options only option one is correct because it is true from both ways.

It can be written as If a shape has exactly four sides and four right angles then the shape is a rectangle.

  • A shape is a trapezoid if and only if the shape has a pair of parallel sides.  

It is not true because a if the shape has a pair of parallel sides then it can be a parallelogram.

  • A shape is a triangle if and only if the shape has three sides and three acute angles.

 It is not true because a if the shape has three sides and three acute angles then it is only a acute triangle.

  • A shape is a square if and only if the shape has exactly four congruent sides.

 It is not true because if the shape has exactly four congruent sides then it can also be rhombus.

So the only true bi-conditional statement is "A shape is a rectangle if and only if the shape has exactly four sides and four right angles. "

Oksana_A [137]3 years ago
4 0

Answer:

The first one: A shape is a rectangle if and only if the shape has exactly four sides and four right angles.

Step-by-step explanation:

The true statement is the first one:

A shape is a rectangle if and only if the shape has exactly four sides and four right angles.

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\text{Probability}=\frac{1}{3}

Step-by-step explanation:

Given : A pair of fair dice is rolled, If the sum of the spots is 7.

To find : Determine the probability that one die showed a 2 ?

Solution :

A pair of fair dice is rolled the outcomes are,

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

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(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

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\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total outcome}}

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