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MAXImum [283]
3 years ago
13

Step 1: Calculate the measures of center for Mrs. Hampton's data in the dot plot (round your answer to the nearest tenths place)

. Show your work and briefly explain each step. (Measures of Center are the Mean and Median of a data set)

Mathematics
1 answer:
Deffense [45]3 years ago
8 0

*dot plot is shown in the attachment below

Answer:

Mean = 6.3

Median = 6

Step-by-step explanation:

Measures of centre, mean and median, can be calculated as follows:

First, bear in mind that each dot represents a value in the data set.

==>Mean:

Mean is the sum of all values in the data set divided by the number of data set we have.

The sum can be calculated as follows:

0 (1) = 0

4 (3) = 12

5(8) = 40

6(3) = 18

7(1) = 7

8(5) = 40

9(2) = 18

10 (3) = 30

Sum = 0+12+40+18+7+40+18+30 = 165

No of data set = 26

Mean = 165/26 = 6.346 ≈ 6.3 (nearest tenth place)

==>Median: this is the middle value in the data set. Since the number of data set is even number (26) , the middle value lies between the 13th and 14th data points. The average of the 13th and 14th data points will give us the median value.

Thus, the 13th and 14th values are both 6.

Therefore, median = (6+6) ÷ 2 = 6

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Step-by-step explanation:

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).
lutik1710 [3]

Answer:

The arc length is \dfrac{21}{16}

Step-by-step explanation:

Given that,

The given curve between the specified points is

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

The points from (\dfrac{9}{16},1) to (\dfrac{9}{8},2)

We need to calculate the value of \dfrac{dx}{dy}

Using given equation

x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}

On differentiating w.r.to y

\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})

\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})

\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})

\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}

We need to calculate the arc length

Using formula of arc length

L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}

Put the value into the formula

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}

L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}

L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}

L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}

L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}

L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}

Put the limits

L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})

L=\dfrac{21}{16}

Hence, The arc length is \dfrac{21}{16}

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Step-by-step explanation:

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Aleksandr-060686 [28]

Answer:

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Step-by-step explanation:

Answer:

The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices (x_1,y_1),(x_2,y_2),(x_3,y_3) is given by the determinant of the following matrix:

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In this question:

Vertices (1,0) (5,0) (3,4). So

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A = \pm 0.5*(20-4)

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