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ycow [4]
3 years ago
14

Help! Geometry Triangle Congruence

Mathematics
1 answer:
Feliz [49]3 years ago
7 0

Answer:

Step-by-step explanation:

Thank your for revising and improving the image.

Assuming DF is a straight line,

Given E is the mid-point of DF

mCE = mGF

CE || GF

Then

mDE = mEF (E midpoint of DF)

Angle DEC = Angle EFG (corresponding angle, DF transversal of parallel lines CE and GF)

mCE = mGF (given)

Then triangles DCE and EGF are congruent by reason SAS (side-angle-side)

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A distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters, are listed below: 67.4
skad [1K]

Answer:

a) \bar X =\frac{67.401+67.400+67.402+67.396+67.406+67.401+67.396+67.401+67.405+67.404}{10}=67.401

b) The sample deviation is calculated from the following formula:

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And for this case after replace the values and with the sample mean already calculated we got:

s=0.0036

If we assume that the data represent a population then the standard deviation would be given by:

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}

And then the deviation would be:

\sigma =0.00319

Step-by-step explanation:

For this case we have the following dataset:

67.401, 67.400, 67.402, 67.396, 67.406, 67.401, 67.396, 67.401, 67.405, and 67.404

Part a: Determine the most probable value.

For this case the most probably value would be the sample mean given by this formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

And if we replace we got:

\bar X =\frac{67.401+67.400+67.402+67.396+67.406+67.401+67.396+67.401+67.405+67.404}{10}=67.401

Part b: Determine the standard deviation

The sample deviation is calculated from the following formula:

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And for this case after replace the values and with the sample mean already calculated we got:

s=0.0036

If we assume that the data represent a population then the standard deviation would be given by:

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}

And then the deviation would be:

\sigma =0.00319

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