Two equivalent ratios or rates

one vertex of a triangle is located at (0,5) on a coordinate grid after a transformation the vertex is located at(5,0) which tra

Alik [6]

Answer:

It's a rotation clockwise of 90 degrees about the origin.

Step-by-step explanation:

The given point moves from y = 5 to x = 5 so it has passed through 90 degrees about the origin.

It's a rotation clockwise of 90 degrees about the origin.

6 0

3 years ago

ANSWER THIS PLZZZZZZ

Citrus2011 [14]

I would say it’s B because out of ten it makes the mean 1 point higher

6 0

3 years ago

A student records the repair cost for 22 randomly selected dryers. A sample mean of $98.78 and standard deviation of $15.49 are

Gelneren [198K]

Answer:

The critical value that should be used is T = 2.0796.

The 95% confidence interval for the mean repair cost for the dryers is between $91.912 and $105.648.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 22 - 1 = 21

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 21 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.0796, which is the critical value that should be used.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.0796\frac{15.49}{\sqrt{22}} = 6.868$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 98.78 - 6.868 = $91.912

The upper end of the interval is the sample mean added to M. So it is 98.78 + 6.868 = $105.648

The 95% confidence interval for the mean repair cost for the dryers is between $91.912 and $105.648.

8 0

3 years ago

~PLEASE HELP ASAP OFFERING 10 POINTS~

Marina CMI [18]

T = c + cb
t - c = cb
(t - c)/c = b
b = (t - c)/c

6 0

3 years ago

¿Para cuál(es) valor(es) de p las rectas de ecuación x − 1/p = 2 − y/p y x − 1/1 − p = y − 2/2 son perpendiculares? A) Solo para

Akimi4 [234]

Respuesta:

C) Solo para el -1

Explicación paso a paso:

Para resolver este problema, debemos de determinar la pendiente en cada una de las ecuaciones provistas:

$\frac{x-2}{p}=\frac{2-y}{p}$

y

$\frac{x-1}{1-p}=\frac{y-2}{2}$

ahora bien, necesitamos conocer el valor de la pendiente de una de las dos ecuaciones. Tomemos la primera ecuación y resolvámosla para y:

$\frac{x-2}{p}=\frac{2-y}{p}$

Multiplicamos ambos lados para p y obtenemos:

x-1=2-y

volteamos la ecuación y nos da:

2-y=x-1

pasamos el 2 a restar al otro lado y nos da:

-y=x-1-2

-y=x-3

y dividimos ambos lados de la ecuación dentro de -1

y=-x+3

esta ecuación ya tiene la forma pendiente intercepto:

y=mx+b

donde m es nuestra pendiente:

$m_{1}=-1$

Esta es la pendiente de una de las dos ecuaciones, para que la segunda ecuación sea perpendicular a la primera, su pendiente debe de ser el recíproco negativo de la pendiente de la primera ecuación, entonces la pendiente de la segunda ecuación debe ser:

$m_{2}=-\frac{1}{m_{1}}$