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Bas_tet [7]
2 years ago
8

50 POINTS!!!!

Mathematics
1 answer:
amid [387]2 years ago
3 0
The answer is 503.39

Hope this helps
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Number of data values: 1007
kogti [31]

Answer:

The arithmetic mean is 10.9

Step-by-step explanation:

In this question, we are tasked with calculating the arithmetic mean using the information given.

The arithmetic mean or otherwise called the average is the sum of all the values in a given set divided by the number of elements or count of the elements in that particular data set

Mathematically, the arithmetic mean can be found by dividing the sum of the data values by the number of data values.

In equation form, we have Arithmetic -mean = \frac{sum of data values}{number of data values}

Inserting the values, we have Arithmetic mean = 10998/1007 = 10.92 which is 10.9 to the nearest tenth

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2 years ago
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Answer:

69

Step-by-step explanation:

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3 years ago
Elija, Emily, Edward, and Esme each drew a conclusion about the figure below.
Anvisha [2.4K]
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Find 30% decrease from 95<br> 68.40<br> 63.20<br> 66.50
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Answer:

66.50

stay home and be safe

7 0
3 years ago
Given: BD bisects ABC. Auxiliary EA is drawn such that AE || BD. Auxiliary BE is an extension of BC
AnnZ [28]
The required proof is given in the table below:

\begin{tabular}{|p{4cm}|p{6cm}|} &#10; Statement & Reason \\ [1ex] &#10;1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\&#10;2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ &#10;3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ &#10;4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\&#10;5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ &#10;6. \angle ABD\cong\angle BAE & 6. Alternate angles&#10;\end{tabular}
\begin{tabular}{|p{4cm}|p{6cm}|}&#10;7. \angle AEB\cong\angle BAE & 7. Transitive property of equality \\&#10;8. \overline{EB}\cong\overline{AB} & 8. From 7. $\Delta ABE$ is isosceles \\&#10;9. EB = AB & 9. De(finition of congruence \\ 10. $\frac{AD}{DC}=\frac{EB}{BC}$ & 10. Triangle proportionality theorem \\&#10;11. $\frac{AD}{DC}=\frac{AB}{BC}$ & 11. Substitution Property of equality \\[1ex] &#10;\end{tabular}&#10;
7 0
3 years ago
Read 2 more answers
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