In which data set is the mean equal to the median?

A coat and a pair of boots are on sale for 20% off. The regular price of the coat is $225. The regular price of the boots is $13

kondaur [170]

Answer:

Part A = $180

Part B = $104

Step-by-step explanation:

Part A:

You do 255÷5=51

51 x 4 = 180

Part B:

You do 130÷5=26

26 x 4 = 104

7 0

2 years ago

Read 2 more answers

The length of a rectangle is 5 meters less than twice the width. If the area of the rectangle is 273 meters, find the dimensions

AleksandrR [38]

The length of the rectangle = 16 meters

The width of the rectangle = 10.5 meters

Step-by-step explanation:

Given that,

The Area of the rectangle is 273.

Step 1 :

Let, the width of the rectangle be 'x'

The length of the rectangle is 2x - 5

Step 2 :

Area of the rectangle = length $\times$ width

273 = (2x - 5) x

273 = 2x^2 - 5x

Step 3 :

The equation formed is 2x^2 - 5x -273 = 0

a= coefficient of x^2 = 2

b= coefficient of x = -5

c = constant = -273

Find the roots of the equation, to find the width.

Step 4 :

Find the roots using factorizing method,

ac = 2 $\times$ -273 = -546 and b= -5

Factorizing -546 as -26 $\times$ 21

The product of -26 $\times$ 21 = -546

The sum of -26 + 21 = -5

Step 5 :

The roots of the equation 2x^2 - 5x -273 = 0 are x= -26/2 and x= 21/2

The values of x are x= -13 and x= 10.5

Since the width cannot be a negative value, neglect x= -13

Step 6 :

Therefore,

The width of the rectangle = 10.5 meters

The length of the rectangle = 2x-5

= 2(10.5) - 5

= 21 - 5

length = 16 meters

5 0

3 years ago

How do i find the vertex from x intercept form

hjlf

You'll have to convert the equation from x-int to standard form (ax² + by + c), and then convert it to vertex form (y = a(x - h)² + k)

3 0

3 years ago

I don't understand this question at all please help if you do

lorasvet [3.4K]

Answer:

Answer is $\frac{4}{3}$

Step-by-step explanation:

To find the interval of x. Use our equations to equal each other.

$-x^2+2x=y,y=0$

$-x^2+2x=0\\-x(x-2)=0\\-x=0 , (x-2)=0\\x=0 ,2$

$\int\limits^2_0 {-x^2+2x} \, dx$

Integrate.

$\frac{-x^3}{3}+x^2\\(\frac{-2^3}{3}+2^2)-[\frac{-0^3}{3}+0^2]\\-\frac{8}{3} +4-0\\-\frac{8}{3}+\frac{12}{3} =4/3$

Using Desmos I have Graphs of both of the equations you have provided. The problem asks us to find the shaded region between those curves/equations.

Proof Check your interval of x.

7 0

2 years ago

suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. use the 68-95-99.7 rule to

Lelechka [254]

152700 is answer. Try to do these problems on paper then you can see where you went wrong . also try to do the actual question

5 0

3 years ago