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2 years ago

Please help ASAP. What is the length os side BC of the triangle

Mathematics

1 answer:

Nuetrik [128]2 years ago

7 0

nswer:

B C = 24

Explanation:

The triangle is isosceles, because (from the diagram) ∠ A B C = ∠ A C B - note how the angle there is marked on both of these. If these angles are equal, this means that the sides

A C = A B

let A C = A B

x +8=2x-1

8 = x − 1

x = 9

We want to find the length of

B C , which is ( 3 x − 3 ) . We know that x = 9

B C =3 x − 3

B C = 3 ( 9 ) −3

B C = 27 − 3

B C = 24

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Answer:

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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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For 59 - 66

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

For 67 - 74

$x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5$

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For 83 - 90

$x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5$

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:

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

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

$Mean = \frac{2547.5}{37}$

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

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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2 years ago

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2 years ago

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