Ask question

Ask question

All categories

3 years ago

13

a large fire department rceived 198 false alarms last month. This was 12% of the total alarms received. How many alarms were rec

eived last month? Will it be 1,650 Alarms?

1 answer:

Mandarinka [93]3 years ago

7 0

Answer:

Yes, it will be 1,650 alarms

Step-by-step explanation:

All you needed to do was divide 198 by 12, since it's 12%, to find the unit rate. That multiplied by 100 is 1,650.

You might be interested in

One and read or both rode bicycles from the park to Main Street the graphs below represent the time and distance for each studen

Harman [31]

Juan did.

The line on his graph takes longer to get to 8 blocks.;-D

7 0

3 years ago

Write a decimal between 0.5 and 0.75.Then write it as a fraction in simplest form and as a percent

Andre45 [30]

You can do anything between .50 and .75. For example .55. .55=55% .55 also equals 11/20 as a fraction

8 0

3 years ago

How to determine the function with the graph?

Roman55 [17]

I think it’s f (x)= 1

3 0

3 years ago

The solution is (7,7)

Georgia [21]

B is the answer I chose but there are the steps so you can see how to solve it

8 0

2 years ago

A right cylinder has a radius of r inches and height of 2r inches.

lys-0071 [83]

Answer:

Part a) The lateral area is $4r^{2} \pi \ in^{2}$

Part b) The area of the two bases together is $2r^{2} \pi\ in^{2}$

Part c) The surface area is $6r^{2} \pi\ in^{2}$

Step-by-step explanation:

we know that

The surface area of a right cylinder is equal to

$SA=LA+2B$

where

LA is the lateral area

B is the area of the base of cylinder

we have

$r=r\ in$

$h=2r\ in$

Part a) Find the lateral area

The lateral area is equal to

$LA=2\pi rh$

substitute the values

$LA=2\pi r(2r)$

$LA=4r^{2} \pi\ in^{2}$

Part b) Find the area of the two bases together

The area of the base B is equal to

$B=r^{2} \pi\ in^{2}$

so

the area of the two bases together is

$2B=2r^{2} \pi\ in^{2}$

Part c) Find the surface area of the cylinder

$SA=LA+2B$

we have

$LA=4r^{2} \pi\ in^{2}$

$2B=2r^{2} \pi\ in^{2}$

substitute

$SA=4r^{2} \pi+2r^{2} \pi=6r^{2} \pi\ in^{2}$

4 0

3 years ago

Read 2 more answers