Ask question

Ask question

All categories

2 years ago

11

What is the endpoint of a ray

1 answer:

Aleksandr-060686 [28]2 years ago

8 0

Answer:

A ray is a portion of a line. It includes one point on the line, the endpoint of the ray, and all the points of the line in only one direction.

The endpoint of a ray is the staring point of the ray.

You might be interested in

Find the solutions to the equation below check all that apply 4x2 -8x +3 =0

notka56 [123]

Answer:

Hello :

4x² - 8x +3 = 0

4 ( x² - 2x +3/4 ) = 0

4 ( (x² -2x +1) -1 + 3/4) = 0

4 ((x-1)² - 1/4 ) =0

(x-1)² - 1/4 = 0

(x-1)² - ( 1/2 )²=0

( x - 1 -1/2 ) ( x - 1 +1/2 ) = 0

( x - 3/2) ( x - 1/2) = 0

x - 3/2 = 0 or x - 1/2 = 0

two solutions : 3/2 ; 1/2

4 0

3 years ago

18 cajas de manzanas y peras fueron llevadas a la escuela. Las peras trajeron 210 kg en 6 cajas. ¿Cuántos kilos de manzanas fuer

Katyanochek1 [597]

Answer:

504 kilos de manzanas fueron llevados a la escuela.

Step-by-step explanation:

Sean $x$ , $y$ la cantidad de cajas de manzanas y peras, respectivamente. De acuerdo con el enunciado, tenemos el siguiente sistema de ecuaciones:

Total de cajas

$x+y = 18$ (1)

Masa de cada caja de peras

$y\cdot m_{P} = 210$ (2)

$y = 6$ (3)

Masa de caja de manzanas

$m_{M} = m_{P}+7$ (4)

De (2) en (3), tenemos que la masa de cada caja de peras es:

$m_{P} = 35\,kg$

Por (4), encontramos que la masa de cada caja de manzanas es:

$m_{M} = 35\,kg+7\,kg$

$m_{M} = 42\,kg$

Finalmente, de (1) tenemos la cantidad de cajas de manzana:

$x = 18-y$

$x = 12$

La cantidad de kilos de manzanas llevada a la escuela se obtiene al multiplicar la cantidad de cajas de manzana por la masa de cada caja:

$M = (12)\cdot (42\,kg)$

$M = 504\,kg$

504 kilos de manzanas fueron llevados a la escuela.

7 0

3 years ago

In a television series there is exactly one crime committed in every episode. It has been observed that the probability that the

Rama09 [41]

Answer:

$P(x \le 3) = 0.2352$

Step-by-step explanation:

Given

$p = 0.35$ --- male committing a crime in an episode

$n = 5$ -- Number of episodes

Required

Determine the probability of male committing a crime at least 3 times

This question illustrates binomial distribution and will be solved using;

$P(x) = ^nC_x * p^x * (1 - p)^{n-x$

So, the required probability is represented as:

$P(x \ge 3)$

And will be calculated using:

$P(x \ge 3) = P(x = 3) + P(x = 4) + P(x = 5)$

$P(x = 3) = ^5C_3 * (0.35)^3 * (1 - 0.35)^{5-3}$

$P(x = 3) = ^5C_3 * (0.35)^3 * (1 - 0.35)^2$

$P(x = 3) = 10 * (0.35)^3 * (0.65)^2$

$P(x = 3) = 0.1811$

$P(x = 4) = ^5C_4 * (0.35)^4 * (1 - 0.35)^{5-4}$

$P(x = 4) = 5 * (0.35)^4 * (1 - 0.35)^1$

$P(x = 4) = 5 * (0.35)^4 * (0.65)$

$P(x = 4) = 0.0488$

$P(x = 5) = ^5C_5 * (0.35)^5 * (1 - 0.35)^{5-5}$

$P(x = 5) = 1 * (0.35)^5 * (1 - 0.35)^0$

$P(x = 5) = 1 * (0.35)^5 * (0.65)^0$

$P(x = 5) = 1 * (0.35)^5 * 1$

$P(x = 5) = 0.0053$

So:

$P(x \ge 3) = P(x = 3) + P(x = 4) + P(x = 5)$

$P(x \le 3) = 0.1811 + 0.0488 + 0.0053$

$P(x \le 3) = 0.2352$

4 0

3 years ago

I need this nowwww plss

Zielflug [23.3K]

I think the answer is 1/23

Please tell me if I'm wrong.

3 0

3 years ago

Which type of visual aid best allow students to show a numerical information?

ryzh [129]

Graphs

°°°°°°°°°°°°°°°°

4 0

3 years ago