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saveliy_v [14]
3 years ago
8

The diameter of a tree trunk is 23 in. what is the circumference?​

Mathematics
1 answer:
krok68 [10]3 years ago
6 0

Answer:

72.2566311

Step-by-step explanation:

Circumference = 2πr

We don't have the radius, but we do know that half of the diameter is the radius, so 23/2 = 11.5. r = 11.5

Now, plug the radius into the equation:

2π(11.5)

Solve:

2π(11.5) = 72.2566311

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The graphs below have the same shape. What is the equation of the red
Citrus2011 [14]

Answer:

g(x)=2-x² OR g(x)=-x²+2

Step-by-step explanation:

Notice that the red parabola doesn't stretch or compress. It only translates 3 units down. This means taking the original equation f(x)=-x²+5 and subtracting 3 units will get us the equation g(x)=-x²+2 which is also the same thing as g(x)=2-x². Looks like some of your answer choices got cut out of the picture so make sure to choose the answer I provided.

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2 years ago
6c to te second power - 5d + 8
mina [271]
Lets see,
6c to the second power is 36c because 6c times 6c is 36 c
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7 0
3 years ago
I need help on this :,)​
masha68 [24]

Answer:

a. Figure B and Figure C

b. Figure C

c. Figure B and Figure C

Step-by-step explanation:

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3 0
3 years ago
To find the standard deviation of the diameter of wooden​ dowels, the manufacturer measures 19 randomly selected dowels and find
loris [4]

Answer:

Option D -  [0.12

Step-by-step explanation:

Given : The manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s=0.16.

To find : The​ 95% confidence interval for the population standard deviation sigma?

Solution :

Number of sample n=19

The degree of freedom is Df=n-1=19-1=18

The standard deviation of the sample is s=0.16

Applying chi-square table to find critical value,

Upper critical value of \chi^2 is UC=\chi(\frac{0.05}{2},18) = 31.5264

Lower critical value of \chi^2 is  

LC=\chi(1-\frac{0.05}{2},18) = 8.2307

Lower limit of the 95% confidence interval for the population variance

L=\frac{(df)\times (s^2)}{UC}

L=\frac{18\times (0.16^2)}{31.5264}

L=\frac{18\times0.0256}{31.5264}

L=\frac{0.4608}{31.5264}

L=0.0146

Upper limit of the 95% confidence interval for the population variance

U=\frac{(df)\times(s^2)}{LC}

U=\frac{18\times (0.16^2)}{8.2307}

U=\frac{18\times0.0256}{8.2307}

U=\frac{0.4608}{8.2307}

U=0.0559

So, The 95% confidence interval for the population variance is [0.0146, 0.0560]

Now, The 95% confidence interval for the population standard deviation is

[\sqrt{0.0146}

[0.1208

or  [0.12

Therefore, Option D is correct.

The 95% confidence interval for the population standard deviation is  [0.12

6 0
2 years ago
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