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Novosadov [1.4K]

3 years ago

8

What is the length of segment AB?

1 answer:

slamgirl [31]3 years ago

3 0

Answer:

$AB = 10 units$

Step-by-step explanation:

The line of segment AB can be calculated using distance formula, $d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$ , to calculate the distance between point A(6, 2) and point B(0, 10).

A(6, 2) can be (x1, y1),

B(0, 10) can be (x2, y2)

$d = \sqrt{(0 - 6)^2 + (10 - 2)^2}$

$d = \sqrt{(-6)^2 + (8)^2}$

$d = \sqrt{36 + 64}$

$d = \sqrt{100}$

$d = 10$

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Step-by-step explanation:

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$\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

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For 67 - 74

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$\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}$

The mean is then calculated as:

$Mean = \frac{\sum fx}{\sum f}$

$Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}$

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This is calculated as:

$s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}$

So, we have:

$s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}$

$s^2 =\frac{652.8}{36}$

$s^2 =18.1$ -- approximated

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