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vodka [1.7K]
3 years ago
7

I am a regular polygon. Each of my exterior angles has a measure of 6 and 2/3 degrees. How many sides do I have?

Mathematics
1 answer:
Alik [6]3 years ago
6 0
A polygon is a triangle it has 3 sides.
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What is the experamental probability that the coin lands on heads?
sammy [17]

Answer:

The experamental probability that the coin lands on head is 50 %

Step-by-step explanation:

Given:

Experiment:

A coin is Toss

Let the Sample Space be 'S' that is total number of outcomes for a coin has been tossed = { Head, Tail }

∴ n ( S ) =  2

Let A be the event of getting a Head on tossing a coin i.e  { Head }

∴ n( A ) = 1

Now,

\textrm{Probability} =\frac{\textrm{outcomes for the experiment}}{\textrm{total number of outcomes}}

Substituting the values we get

P(A) = \frac{n(A)}{n(S)} \\\\P(A) = \frac{1}{2} \\\\P(A) = 0.5\\\\P(A) = 50\%

The experamental probability that the coin lands on head is 50 %

5 0
3 years ago
Given that log 2 = 0.3010 and log 3 = 0.4771 , how can we find log 6 ? ​
bearhunter [10]

Answer:

\sf \log_{10}6=0.7781

Step-by-step explanation:

<u>Given</u>:

\sf \log_{10} 2 = 0.3010

\sf \log_{10} 3 = 0.4771

To find log₁₀ 6, first rewrite 6 as 3 · 2:

\sf \implies \log_{10}6=\log_{10}(3 \cdot 2)

\textsf{Apply the log product law}: \quad \log_axy=\log_ax + \log_ay

\implies \sf \log_{10}(3 \cdot 2)=\log_{10}3+\log_{10}2

Substituting the given values for log₁₀ 3 and log₁₀ 2:

\begin{aligned} \sf \implies \log_{10}3+\log_{10}2 & = \sf 0.4771+0.3010\\ & = \sf 0.7781 \end{aligned}

Therefore:

\sf \log_{10}6=0.7781

Learn more about logarithm laws here:

brainly.com/question/27953288

brainly.com/question/27963321

5 0
1 year ago
Read 2 more answers
The scale factor of the blueprint of a gymnasium to the actual gymnasium is 1in/15ft. The area of the flow on the blueprint is 1
stira [4]
1/15=144/x
Cross multiply
X=2,160

Gymnasium=2,160 ft2
3 0
3 years ago
Read 2 more answers
Melinda rode her bike 54/100 mile to the library. Then she rode 4/10 mile to the store. How far did Melinda ride her bike in all
weqwewe [10]
0.94 
...............

hope this helps
3 0
3 years ago
Read 2 more answers
A) how many ways are there to choose 10 players to take the field?
Kruka [31]

A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.

B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4. (13−10)! 3!

C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =

10! 3!

9!1! 1!2! 10! 3!

8!2! 2!1! 10! 3!

7!3! 3!0!

= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =

= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =

= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285 1·2·3

<span>possible choices.</span><span> 
</span>

  


5 0
3 years ago
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