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Verizon [17]
3 years ago
14

In a circle, an arc length of 6.6 is intercepted by central angle of 2/3 radians. Determine the length of the radius.

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
3 0

Answer:

9.9

Step-by-step explanation:

To solve this problem, we first need to know that the whole length of the circunference is related to a central angle of 2pi radians. Then, we can solve using a rule of three to find the radius:

arc of 6.6 -> central angle of 2/3

arc of 2*pi*r -> central angle of 2pi

2*pi*r * (2/3) = 6.6 * 2*pi

r * (2/3) = 6.6

r = 6.6 / (2/3) = 9.9

So the length of the radius is 9.9

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D. All of the above.

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2 years ago
Factor out the coefficient of the variable -3.6m + 10.8
MA_775_DIABLO [31]

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Step-by-step explanation: First, factor out 3.6. You should get -m+3. Since the coefficient is just the number being multiplied to the variable (which is m here), your coefficient should be -1 because m is being multiplied to -1 to make -m. Hope this helps!

3 0
3 years ago
How do you solve 8x-7= 3x 9?
gtnhenbr [62]
8x - 7 = 3x + 9
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3 0
2 years ago
PROBABILITY PROBLEM: Can anyone check my answers for this?
Julli [10]
A: it would be 2 fives outta 8 then he picked one more bill out so chances are 4 fives outta 16. making it 1/4 of a probability.
7 0
3 years ago
The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
5 0
3 years ago
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