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3 years ago

What is the area of a circle with a radius of 6 ft? Use 3.14 for π. Round to the nearest tenth if necessary.

Mathematics

2 answers:

Ne4ueva [31]3 years ago

4 0

Step-by-step explanation:

write what you know

A = pi x r^2

r = 6ft

Annd plug and chug

Kamila [148]3 years ago

3 0

Answer:

$\boxed {\boxed {\sf a \approx 113.0 \ ft^2}}$

Step-by-step explanation:

The area of a circle is the space inside the shape. The formula is:

$a= \pi r^2$

We know the radius of the circle is 6 feet and we are using 3.14 for pi.

r= 6 ft
π= 3.14

Substitute the values into the formula.

$a= (3.14) \times (6 \ ft)^2$

Solve the exponent.

(6 ft)² = 6 ft * 6 ft = 36 ft²

$a= (3.14) \times 36 \ ft^2$

Multiply.

$a=113.04 \ ft^2$

Let's round to the nearest tenth. The 4 in the hundredth place tells us to leave the 0 in the tenths place.

$a \approx 113.0 \ ft^2$

The area of the circle is approximately <u>113.0 square feet.</u>

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B. Find the standard deviation of the data.

Ivahew [28]

Answer:

b. The standard deviation for class 1 is approximately 18.81415

The standard deviation for class 2 is approximately 16.1527

The standard deviation for class 3 is approximately 12.705

c. The mean of the data indicates that the average scores of Class 1 and Class 2 are equal and less than the average score of Class 3

The standard deviation indicates that the data in Class 1 varies more than the data in Class 2, and also, that the data of Class 2 varies more than the data of Class 3.

Step-by-step explanation:

b. The mean and the standard deviation can be calculated using Microsoft Excel with the values presented here as results as follows;

The standard deviation, σ is given by the formula $\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}}$ , while the formula mean, μ is $\mu = \dfrac{\Sigma x}{N}$

Where;

N = The number of terms = 12

∑x = 946 for Class 1, 953 for Class 2, 953 for Class 3

The mean for Class 1 = ∑x for Class 1/N = 946/12 = 78.8 $\bar 3$

The mean for Class 2 = ∑x for Class 2/N = 953/12 = 79.41 $\bar 6$

The mean for Class 3 = ∑x for Class 3/N = 953/12 = 79.41 $\bar 6$

$\sum \left (x_i-\mu \right )^{2} }$ for Class 1 = 4247. $\bar 6$

$\sum \left (x_i-\mu \right )^{2} }$ /N = for Class 1 = 353.97 $\bar 2$

The standard deviation for class 1 σ₁, is given as follows;

$\sigma_1 =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}} = \sqrt{353.97\bar 2 } \approx 18.81415$

The standard deviation for class 1 = 18.81415

$\sum \left (x_i-\mu \right )^{2} }$ for Class 2 = 3130.9167

$\sum \left (x_i-\mu \right )^{2} }$ /N = for Class 2 = 260.90972

The standard deviation for class 2 σ₂, is given as follows;

$\sigma_2 =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}} \approx \sqrt{260.90972 } \approx 16.1527$

The standard deviation for class 2 = 16.1527

$\sum \left (x_i-\mu \right )^{2} }$ for Class 3 ≈ 1936.917

$\sum \left (x_i-\mu \right )^{2} }$ /N = for Class 3 ≈ 161.4097222

The standard deviation for class 3 σ₃, is given as follows;

$\sigma_3 =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}} \approx \sqrt{161.40972 } \approx 12.705$

The standard deviation for class 3 ≈ 12.705

c. The mean indicates that the average score for Class 1 is higher than the average score for Class 2, and that the average score for Class 2 is the same as the average score for Class 3

However, the standard deviation indicates that the variability of the scores for Class 1 is higher than the variability of the scores for Class 2, and that there is also a higher variability in the scores of Class 2 than the variability in the scores of class 3

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