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S_A_V [24]
2 years ago
8

The circumference of a circle is 6π m. What is the area, in square meters? Express your answer in terms of \piπ.

Mathematics
1 answer:
zaharov [31]2 years ago
5 0

Answer:

Area = 9π sq. m.

Step-by-step explanation:

C = πd

6π = πd

d = 6

r = 3

A = πr^{2} = π3^{2} =  9π sq. m.

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Answer:

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Step-by-step explanation:

1/4+1/10 is 7/20 or 0.35 in decimal form. If they both together scored 21 points then you can add another 7/20 (21 points) and have 14/20 (42 points). You have 6/20 more left. If you can solve what 1/20 could be then you would find the answer. Does this helps?

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Find the sum and express it in simplest form <br> (6y-2c+2)+(-3y+4c)
Taya2010 [7]
Well, kid all you have to do is add like terms and place the subjected terms in alphabetical order. 
1. (6y-2c+2)+(-3y+4c)
2. 6y + - 3y = 3y
3. -2c+4c =2c
4. the positive in the first set of parenthesis has to term other than its number by itself. (so it remains alone and only positive 2)
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3 years ago
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Annette [7]
Hope this helps you.

8 0
3 years ago
The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co
Alona [7]

Answer:

95% Confidence interval for the variance:

3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:

1.9108\leq \sigma \leq 5.8819

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

s^2=2.89^2=8.3521

The confidence interval for the variance is:

\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128

Then, the confidence interval can be calculated as:

\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819

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