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Lostsunrise [7]
3 years ago
6

I need help asap what is the mode of 63, 70, 75, 80, 85, 85, 85, 92, 92, 95, 99

Mathematics
2 answers:
Bad White [126]3 years ago
8 0
<h3>MODE</h3>

Mode is the number that appears the most often.

There are 8 numbers. Here is the amount each number appears:

63 \rightarrow 1\ time \\ 70 \rightarrow 1\ time \\ 75 \rightarrow 1\ time \\ 80 \rightarrow 1\ time \\ 85 \rightarrow 3\ times \\ 92 \rightarrow 2\ times \\ 95 \rightarrow 1\ time \\ 99 \rightarrow 1\ time

Since 85 appears the most, 85 is the mode

Ann [662]3 years ago
4 0
The mode would be 85 since it occurs the most out of the whole data set.
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Given: AB || DE , AD bisects BE.<br> Prove: ABC = DEC using the ASA postulate.
Ket [755]

Answer:

As per ASA postulate, the two triangles are congruent.

Step-by-step explanation:

We are given two triangles:

\triangle ABC and \triangle DEC.

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AB || DE.

Let us have a look at two properties.

1. When two lines are parallel and a line intersects both of them, then <em>alternate angles </em>are equal.

i.e. AB || ED and \angle B and \angle E are alternate angles \Rightarrow \angle B = \angle E.

2. When two lines are cutting each other, angles formed at the crossing of two, are known as <em>Vertically opposite angles </em>and they are are <em>equal</em>.

\Rightarrow \angle ACB = \angle DCE

Also, it is given that <em>AD bisects BE</em>.

i.e. EC = CB

1. \angle B = \angle E

2. EC = CB

3. \angle ACB = \angle DCE

So, we can in see that in \triangle ABC and \triangle DEC, two angles are equal and side between them is also equal to each other.

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8 0
3 years ago
Which strategy should be used to correctly solve this problem?
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the best way to do it is actually not listed
to have max area with minimu perimiter, try to get the sides as close measure to each other as possible


I would pick B though

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