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3 years ago

When is a rectangle a square?

Mathematics

2 answers:

ololo11 [35]3 years ago

6 0

Answer:

I think option B i.e. when its angles are right angles.

tankabanditka [31]3 years ago

6 0

Answer:

A. When its sides are congruent

Step-by-step explanation:

By definition a square is a four sided polygon (shape) which has four right angles and four sides of equal length.

So, when a rectangle has all congruent sides, its a square because its a rectangle so it has four right angles, and congruent sides are sides of the same length.

B is incorrect because a rectangle can have four right angles but have differing side lengths

C is incorrect because again a rectangle can have parallel sides but have differing side lengths

D is incorrect because it can have convex angles but again have differing side lengths

You might be interested in

The following triangles are similar. Find side AB.

tankabanditka [31]

Answer:

Step-by-step explanation:

Set up ratios of corresponding sides. x+2 is to 6 as 2 is to 3:

$\frac{x+2}{6}=\frac{2}{3}$

Now cross multiply to get

3(x + 2) = 12 and

3x + 6 = 12 so

3x = 6 and

x = 2. That means that side AB, x + 2, is 2 + 2 which is 4

4 0

3 years ago

Find the L.C.M of 6,9

Oxana [17]

Answer:

Lcm=18

Step-by-step explanation:

First do prime factorization of 6 and 9

common factor of 6 is 3

and remaining factor is 2×3

Now,

C.F×R.F= 3×2×3

=18

6 0

2 years ago

A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li

svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

15 drivers are randomly selected, so $n = 15$ .

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year. This means that $\pi = 0.08$ .

What is the probability of getting 3 or more who were involved in a car accident last year?

This is $P(X \geq 3)$ .

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$