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mart [117]
3 years ago
15

A circle has a radius of 10. An arc in this circle has a central angle of 72 degrees. What is the length of the arc? btw, the ar

c is aligned with the radius
Mathematics
1 answer:
yaroslaw [1]3 years ago
8 0

Answer:

4 Pi

Explanation:

72/360 = 1/5 so the length of the arc is (1/5) the circumference or

(1/5) ( 2 Pi * 10)  => C = 2 Pi r

(20/5) Pi =

4 Pi

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Two functions are shown below.<br> f(x)=4^x-62<br> g(x)=x^2-2x+5<br><br> evaluate g(-2)+f(3)
algol [13]

Answer:

...

Step-by-step explanation:

5 0
2 years ago
What is the probability that more than twelve loads occur during a 4-year period? (Round your answer to three decimal places.)
Nataly_w [17]

Answer:

Given that an article suggests

that a Poisson process can be used to represent the occurrence of

structural loads over time. Suppose the mean time between occurrences of

loads is 0.4 year. a). How many loads can be expected to occur during a 4-year period? b). What is the probability that more than 11 loads occur during a

4-year period? c). How long must a time period be so that the probability of no loads

occurring during that period is at most 0.3?Part A:The number of loads that can be expected to occur during a 4-year period is given by:Part B:The expected value of the number of loads to occur during the 4-year period is 10 loads.This means that the mean is 10.The probability of a poisson distribution is given by where: k = 0, 1, 2, . . ., 11 and λ = 10.The probability that more than 11 loads occur during a

4-year period is given by:1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]= 1 - [0.000045 + 0.000454 + 0.002270 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.125110+ 0.125110 + 0.113736]= 1 - 0.571665 = 0.428335 Therefore, the probability that more than eleven loads occur during a 4-year period is 0.4283Part C:The time period that must be so that the probability of no loads occurring during that period is at most 0.3 is obtained from the equation:Therefore, the time period that must be so that the probability of no loads

occurring during that period is at most 0.3 is given by: 3.3 years

Step-by-step explanation:

8 0
4 years ago
The numbers​ 1, 2,​ 3, 4, and 5 are written on slips of​ paper, and 2 slips are drawn at random one at a time without replacemen
mamaluj [8]
<h2>Answer:</h2>

(a)

The probability is :  1/2

(b)

The probability is :  1/2

<h2>Step-by-step explanation:</h2>

The numbers​ 1, 2,​ 3, 4, and 5 are written on slips of​ paper, and 2 slips are drawn at random one at a time without replacement.

The total combinations that are possible are:

(1,2)   (1,3)    (1,4)    (1,5)

(2,1)   (2,3)   (2,4)   (2,5)

(3,1)   (3,2)   (3,4)   (3,5)

(4,1)   (4,2)   (4,3)   (4,5)

(5,1)   (5,2)   (5,3)   (5,4)

i.e. the total outcomes are : 20

(a)

Let A denote the event that the first number is 4.

and B denote the event that the sum is: 9.

Let P denote the probability of an event.

We are asked to find:

               P(A|B)

We know that it could be calculated by using the formula:

P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}

Hence, based on the data we have:

P(A\bigcap B)=\dfrac{1}{20}

( Since, out of a total of 20 outcomes there is just one outcome which comes in A∩B and it is:  (4,5) )

and

P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 9

(4,5) and (5,4) )

Hence, we have:

P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

(b)

Let A denote the event that the first number is 3.

and B denote the event that the sum is: 8.

Let P denote the probability of an event.

We are asked to find:

               P(A|B)

Hence, based on the data we have:

P(A\bigcap B)=\dfrac{1}{20}

( since, the only outcome out of 20 outcomes is:  (3,5) )

and

P(B)=\dfrac{2}{20}

( since, there are just two outcomes such that the sum is: 8

(3,5) and (5,3) )

Hence, we have:

P(A|B)=\dfrac{\dfrac{1}{20}}{\dfrac{2}{20}}\\\\i.e.\\\\P(A|B)=\dfrac{1}{2}

7 0
3 years ago
Number 8 only please
givi [52]

Answer:

Should be roughly 79.76% if I'm doing this right.

Step-by-step explanation:

Looking at the problem, you're given 3 things:

1) 6.8 million tonnes are recycled;

2) 6.8 million tonnes is 27% of the total household waste;

3) 20.1 million tonnes could be recycled.

Since they're asking for what percentage of <em>total</em> household waste could be recycled, you're looking at a multi-step cross multiplication problem.

It's not very difficult though, don't worry.

So I'm gonna start by telling you to just forget that the word "million" exists for the moment. You only need 6.8, 20.1, and 27%.

Setting up the first part of the problem requires you to convert 27% into a fraction. That's \frac{27}{100}.

6.8 is 27% of some mysterious number that we need to figure out what percent of household waste could be recycled, so that'll be set up on the opposite side of the equal sign as the equivalent to 27.

So we're starting out with this: \frac{6.8}{?} =\frac{27}{100}

Since you're trying to initially find the total household waste, we're setting x as the variable to solve for.

\frac{6.8}{x} =\frac{27}{100}

Cross-multiplying** means you multiply the numerator of one side of the equal sign to the denominator of the other side of the equal sign, and vice versa (denominator of one side to numerator of other). So multiply 6.8 to 100, and 27 to x.

680=27x

Now that you have this, you have to isolate x. Divide both sides by 27 to do this.

25.185185...=x

I'm gonna round that for simplicity's sake, because 185 repeats.

25.2=x

So we have 25.2 million tonnes of total household waste. Great! Now we can find the <em>percentage </em>of household waste that could be recycled.

Since we have how much could be recycled and the total, we just need to solve for the percent this time. So this is what the set-up will look like:

\frac{20.1}{25.2} =\frac{x}{100}

x being the percent we're solving for.

Cross-multiply again:

2010=25.2x

Divide both sides by 25.2:

79.761904761904...=x

Then round for simplicity's sake:

79.76=x

And you get 79.76% of household waste could be recycled.

Hope this helped!

**Be careful with cross-multiplying when you get to higher level math though. Not every situation allows for cross-multiplying. In fact, you can only cross-multiply when there's an equal sign.

6 0
3 years ago
Which equation is equivalent to this formula?
lana [24]
The equation equivalent to this formula would be D)b1=2A-b2
4 0
4 years ago
Read 2 more answers
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