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2 years ago

15

Robert can complete 10 inspections in 8 hours. Which represents the unit rate for the number of inspections he completes per hou

r?
0.8 of an inspection in 1 hour
4 inspections in 5 hours
1.25 inspections in 1 hour
5 inspections in 4 hours

1 answer:

a_sh-v [17]2 years ago

8 0

Answer:

1.25 inspections in 1 hour

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company loses $5,400 as the result of a manufacturing defect. Each of the 8 owners have agreed to pay an equal amount, x, to pay

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Answer:

Step-by-step explanation:

5400 / 8 = x

675 = x

Each owner pays 675

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1 year ago

45% as fraction in the simplest form

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Answer:9/20

Step-by-step explanation: 45/100=9/20, 45÷5=9, 100÷5=20 both the denominater and the numerator are divisible by 5.

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2 years ago

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A ballroom in a hotel has the shape shown. The manager needs to determine how much carpet is needed to recarpet this area. All a

Mars2501 [29]

Area of a rectangle=length x width

You have two rectangles:

First rectangle (the biggest):
length=9 yd
width=12 yd-6 yd=6 yd

Area of the biggest rectangle=(9 yd)(6 yd)=54 yd²

Second rectangle (the smallest)
length=6 yd
width=3 yd

Area of the smallest rectangle=(6 yd)(3 yd)=18 yd²

Area of the ballroom=area of the biggest rectangle + area of the smallest rectangle.

area of the ballroom=54 yd² +18 yd²=72 yd²

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2 years ago

The description for a certain brand of house paint claims a coverage of 460 ft2/gal. (a) express this quantity in square meters

Angelina_Jolie [31]

Given 460 $ft^2/gal$

a) convert it into $m^2/l$

=> It means $ft^2 -> m^2$ and $gal -> l$

1 ft = 0.3048 m

$1 ft^2 = 0.092903 m^2$

1 gal = 3.78541 l

$460ft^2= [tex]460\frac{ft^2}{gal} = \frac{42.7354m^2}{3.78541l} = 11.2895m^2/l$ m^2[/tex].

b. $11.2895\frac{m^{2} }{l} *1000\frac{l}{m^3} =11289.5 m^{-1}$

c.Given= $460\frac{ft^2}{gal}$

To find inverse, divide it by 1.

Inverse- $\frac{1}{460} \frac{gal}{ft^2}$

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3 years ago

En un aeropuerto dos aviones A1 y A1 se acercan para aterrizar, si la ecuación de la trayectoria del primero es -x+2y-100=0, mie

liraira [26]

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

-------------------------

La trayectoria del primero es:

$-x + 2y - 100 = 0$

$x_1 = 2y - 100$

Para el segundo, tiene-se que:

$-x + 200 + 2y = 0$

$x_2 = 2y + 200$

Igualando los valores de x:

$x_1 = x_2$

$2y - 100 = 2y + 200$

$0y = 300$

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

Un problema similar es dado en brainly.com/question/24653364

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2 years ago