The three angles of a triangle are in the ratio 3:3:4. What information can you give about the triangle?

2 answers:

7 0

The information tells us that this triangle is an isosceles triangle meaning two sides and two angles are congruent. Also, this is not a right triangle as the only isosceles right triangle has a ratio of 1:1:2

Hope this helps!

Vera_Pavlovna [14]3 years ago

4 0

Let the angles be x.
3x + 4x + 4x = 180 (angle sum property of a triangle is 180 degrees.)
12x = 180
x = 180/12
x= 15
the angles are 45, 60 and 60
it is an isosceles triangle.

You might be interested in

Is the Mandelbrot set locally connected?

MariettaO [177]

Answer:

Step-by-step explanation:

It is conjectured that the Mandelbrot set is locally connected. This famous conjecture is known as MLC (for Mandelbrot locally connected). By the work of Adrien Douady and John H. Hubbard, this conjecture would result in a simple abstract "pinched disk" model of the Mandelbrot set. In particular, it would imply the important hyperbolicity conjecture mentioned above.

The work of Jean-Christophe Yoccoz established local connectivity of the Mandelbrot set at all finitely renormalizable parameters; that is, roughly speaking those contained only in finitely many small Mandelbrot copies.[19] Since then, local connectivity has been proved at many other points of {\displaystyle M}M, but the full conjecture is still open.

5 0

3 years ago

A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n indepe

-Dominant- [34]

Answer: 0.0031

Step-by-step explanation:

Binomial distribution formula :-

$P(x)=^nC_xp^x(1-p)^{n-x}$ , where P(x) is the probability of x successes in the n independent trials of the experiment and p is the probability of success.

Given : A binomial probability experiment is conducted with the given parameters.

$n=9,\ p=0.8,\ x\leq3$

Now, $P(x\leq3)=P(3)+P(2)+P(1)+P(0)$

$=^9C_3(0.8)^3(1-0.8)^{9-3}+^9C_2(0.8)^2(1-0.8)^{9-2}+^9C_1(0.8)^1(1-0.8)^{9-1}+^9C_0(0.8)^0(1-0.8)^9\\\\=\dfrac{9!}{3!6!}(0.8)^3(0.2)^6+\dfrac{9!}{2!7!}(0.8)^2(0.2)^7+\dfrac{9!}{1!8!}(0.8)(0.2)^8+\dfrac{9!}{0!9!}(0.2)^9=0.003066368\approx0.0031$