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2 years ago

May I please receive help?

Mathematics

1 answer:

romanna [79]2 years ago

3 0

Answer:

$\frac{10}{9}\pi$ yards

Step-by-step explanation:

The circumference of this circle = 2πr

= 8π

We have to find the arc length of 50°, or 50/360 times the circumference of the circle(because there is 360° in a circle)

$\frac{50}{360} * 8\pi = \\\\\frac{400\pi}{360} =\\\\ \frac{10\pi}{9}$

-Chetan K

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Answer:

$Mean = 68.9$

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Step-by-step explanation:

Given

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

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$x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5$

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For 83 - 90

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So, the table becomes:

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

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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