Donna wrapped 45 gifts. Donna wrapped 9 times as many gifts as Hong. Let be the number of gifts that Hong wrapped.

cestrela7 [59]

Answer:

d.

Step-by-step explanation:

$\bigg( \frac{16}{49} \bigg)^{ - \frac{3}{2} } \\ \\ = \bigg( \frac{49}{16} \bigg)^{ \frac{3}{2} } \\ \\ = \bigg( \frac{ {7}^{2} }{ {4}^{2} } \bigg)^{ \frac{3}{2} } \\ \\ = \bigg( \frac{ {7}}{ {4} } \bigg)^{ \frac{2 \times 3}{2} } \\ \\ = \bigg( \frac{ {7}}{ {4} } \bigg)^{ 3} \\ \\ = \frac{343}{64} \\$

3 0

3 years ago

A survey found that the median number of calories consumed per day in a certain country was 3,304 and the mean was 3,204.9 calor

insens350 [35]

Answer:

Skewed to the left

Step-by-step explanation:

Given

$Median = 3304$

$Mean = 3204.9$

Required

The type of distribution

From the given data, we have:

$Median \ne Mean$ --- Mean and Median are not equal

and

$Median > Mean$ --- Median is greater than mean

<em>When the median is greater than the mean; the histogram is expected to be left skewed</em>

5 0

3 years ago

PLEASEEE HELP MEEEE<br><br>NO LINKSSS

Artemon [7]

Answer:

$\huge\boxed{(2,\ 3);\ (-1,\ 6)}$

Step-by-step explanation:

$f(x)=x^2-2x+3\to y=x^2-2x+3\\f(x)=-x+5\to y=-x+5\\\\\left\{\begin{array}{ccc}y=x^2-2x+3&(1)\\y=-x+5&(2)\end{array}\right$

substitute (1) to (2):

$x^2-2x+3=-x+5$ add x to both sides

$x^2-2x+x+3=-x+x+5$

$x^2-x+3=5$ subtract 5 from both sides

$x^2-x+3-5=5-5$

$x^2-x-2=0\\\\x^2+x-2x-2=0\\\\x(x+1)-2(x+1)=0\\\\(x+1)(x-2)=0\iff x+1=0\ \vee\ x-2=0\\\\\boxed{x=-1\ \vee\ x=2}$

substitute the values of x to (2):

$x=-1\\\\y=-(-1)+5=1+5=6\\\\\huge\boxed{(-1,\ 6)}\\\\x=2\\\\y=-2+5=3\\\\\huge\boxed{(2,\ 3)}$

3 0

3 years ago

Escribe una ecuación de la recta que pasa por el punto (5, –8) con pendiente 5.

Nana76 [90]

Considerando la expresión de una ecuación lineal, la ecuación de la recta que pasa por el punto (5, –8) con pendiente 5 es y= 5x - 33.

<h3>Ecuación lineal</h3>

Una ecuación lineal o línea se puede expresar en la forma y = mx + b

donde

x y son coordenadas de un punto.
m es la pendiente.
b es la ordenada al origen y representa la coordenada del punto donde la línea cruza el eje y.

La ecuación lineal se puede expresar también mediante la ecuación punto-pendiente de la recta, que se plantea si se conoce la pendiente de la recta y cualquiera de sus puntos. Con ello queda determinada la recta, conociendo la pendiente “m” y un punto (x1, y1) dado:

y - y1= m(x -x1)

<h3>Ecuación de la recta en este caso</h3>

En este caso, la recta que pasa por el punto (5, –8) con pendiente 5.

Reemplazando en la ecuación punto-pendiente de la recta:

y - (-8)= 5(x -5)

Resolviendo:

y + 8= 5(x -5)

Aplicando propiedad distributiva:

y + 8= 5x - 5×5

y + 8= 5x - 25

Aislando la variable "y":

y= 5x - 25 - 8

Finalmente, la ecuación de la recta que pasa por el punto (5, –8) con pendiente 5 es y= 5x - 33.

Aprende más sobre ecuación de una recta:

brainly.com/question/25243069

brainly.com/question/19260315

brainly.com/question/24766917

#SPJ1

4 0

2 years ago

PLEASE HELP!!!! WILL MARK BRAINLIEST!!!!

irga5000 [103]

Answer:

None of these.

Step-by-step explanation:

Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.

Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.

In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.

If we look at the first term x^3 and we divide it by 13 we get $\frac{x^3}{13}$ we cannot reduce it so it is not a fraction so 13 is not a factor of P

None of these is the right option.

4 0

3 years ago

Read 2 more answers