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7nadin3 [17]
3 years ago
5

Find the area of a circle with a circumference of 37.67 units

Mathematics
2 answers:
liberstina [14]3 years ago
5 0

The circumference of a circle is pi x diameter. Therefore, we can use this in reverse order as well.

I’ll be using 3.14 as pi, but if you need a more exact answer you can always use the pi function on a calculator.

37.68/3.14 = 12

Therefore, the diameter must be 12.

Since the radius is half the diameter, we can divide by 2 to find the radius.

12/2 = 6

Therefore, the radius of our circle is 6.

The area of a circle is equal to pi x radius squared.

Since we know our radius, we can solve.

Using PEMDAS, we know to use exponents first, so 6 squared is equal to 36.

From there, we can multiply.

36 x 3.14 = 113.04

And there we have your answer! 113.04 is the area of the circle.

Hope this helped :)

ankoles [38]3 years ago
4 0

Answer:

112.91 units squared

Step-by-step explanation:

The formula for the circumference is \pi* diametre

The formula for the area of the circle is \pi *radius^{2}

To work this out you would first divide the circumference of 37.68 by pi to work out the diameter,which gives you 11.99. This is because the formula for the circumference is pi * diameter and so by dividing by pi it will leave you with the diameter. Then you would divide the diameter of 11.99 by 2, which gives you 5.995. This is because now you are left with the radius. Then you would multiply pi by the radius of 5.995 squared, which gives you 112.91.

1) Divide 37.68 by pi.

37.68/\pi =11.99

2) Divide 11.99 by 2.

11.99/2=5.995

3) Multiply pi by 5.995 squared.

\pi *5.995^{2} =112.91

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part b is: if the energy release of one earthquake is 10,000 times that of another, how much larger is the Richter scale reading
kotegsom [21]

You could use the information from part A to get B. I'm not sure if you want A or not, so I'll do it as well.

A

Eo = 10^4.4 Joules

E = 2 * 10^15

Formula

M = (2/3) log (E/Eo)

M = 2/3 * log (2 * 10^15/10^(4.4) )

M = 2/3 * log( 7.9621* 10^10)

M = 2/3 * 10.901

M = 7.26735 on the Richter scale. That is a huge amount of energy.

Part B

Suppose that you use Eo and your base. Eo is 10^4.4

Now the new earthquake is E = 10000 * Eo

So what you get now is M = (2/3)* Log(10000 * Eo / Eo )

The Eo's cancel out.

M = 2/3 * log(10000)

M = 2/3 of 4

M = 8/3

M = 2.6667 difference in the Richter Scale Reading. It is still an awful lot of energy.

What this tells you is that if the original reading was (say) 6 then the 10000 times bigger reading would 8.266667

Answer: M = 2.6667


8 0
3 years ago
Read 2 more answers
3p + 4r = q solve for p
Aleksandr-060686 [28]

Answer:

p = \dfrac{q - 4r}{3}

Step-by-step explanation:

3p + 4r = q

3p = q - 4r

p = \dfrac{q - 4r}{3}

7 0
4 years ago
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An oblique hexagonal prism has a base area of 42 square cm. The prism is 4 cm tall and has an edge length of 5 cm. An oblique he
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Answer:

78

Step-by-step explanation:

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1.) Use a property to write an equivalent expression for 12*(100-5). Which property did you use?
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1. 12*100 - 12*5

You would use the distributive property to get this equation (it is equivalent to the original one)

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3 0
3 years ago
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2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi
Oksana_A [137]

Answer:

(a) E(X) = 950

(b) $ COV = 0.007255$

(c) P(X > 980) = 0.00001\\\\

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$ COV = \frac{\sigma}{E(X)} $

Where the standard deviation is given by

\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892

So the coefficient of variance is

$ COV = \frac{6.892}{950} $

$ COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

P(X > 980)  = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980)  = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980)  = 1 - P(Z < 4.28)\\\\

The z-score corresponding to 4.28 is 0.99999

P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0
3 years ago
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