All categories

3 years ago

Find the circumference of the circle. Use pi = 3.14

Mathematics

2 answers:

svlad2 [7]3 years ago

8 0

B. 18.84

Circumference = 2 π r

r is the radius of the circle
π is the mathematical constant Pi

2(3.14) x (3) =

6.28 x 3 =

18.84

I hope I helped! Thanks for posting your question! :)

kotegsom [21]3 years ago

5 0

Answer:

the answer is the option B

$18.84\ m$

Step-by-step explanation:

we know that

The circumference of a circle is equal to

$C=2\pi r$

In this problem we have

$r=3\ m$

$\pi=3.14$

substitute the values

$C=2(3.14)(3)=18.84\ m$

You might be interested in

Tim and Stacy are volunteers who are helping with the town marathon. They're expecting 260 participants, but only 156 has have r

Amiraneli [1.4K]

Answer:

45%

Step-by-step explanation:

hope it helps,

5 0

2 years ago

Read 2 more answers

The parking rates at the hospital’s parking garage are 50 cents for the first hour and 25 cents for each additional hour. If Jua

Bad White [126]

The total charge for Juan in the parking spot would be $2.25.

Since we know one of those 8 hours is 50 cents, we can multiply .25 by 7 which would give us 1.75. Then we know that the first hour, as stated above, is going to be .50, so we can add that to our total which is 2.25.

Hope this helps!

3 0

2 years ago

Can you please help me out

Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

$\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}$

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

$\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}$ $\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}$

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

$\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}$

Hence, the best option is G.

5 0

11 months ago

What Is the least common multiple of two numbers when one number is a multiple of the other give an example

Tems11 [23]

Like 9 and 9? 9 is a multiple of 3 and 9 time 9 is 81

6 0

2 years ago

Read 2 more answers

Last question! I need help with showing my work! I already have the answer in the attached image below, thanks!

Rainbow [258]

Answer: